1 of 12

STATION 1

STATION 2

Excess aluminum metal reacts with 25.0 g of oxygen gas to produce solid aluminum oxide. Assume the oxygen gas reacts to completion.

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4 Al(s) + 3 O2(g) → 2 Al2O3(s)

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  • Create a BCA table

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  • What is the mass of aluminum oxide produced?

Butane gas combusts completely with 45.7 g of oxygen gas. Carbon dioxide and water are produced. Assume the oxygen gas also reacts to completion.

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2 C4H10(g) + 13 O2(g) → 8 CO2(g) + 10 H2O(g)

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  • Create a BCA table

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  • How many grams of butane reacted?

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  • Calculate the molecules of water that were produced in this reaction.

2 of 12

STATION 1 KEY

STATION 2 KEY

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  • 25.0 g O2 ( 1 mol O2 ) = 0.78125 mol O2

32.00 g O2

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  • 0.52083 mol Al2O3 ( 101.96 g Al2O3 ) =

1 mol Al2O3

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53.10516 → 53.1 g Al2O3

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  • 45.7 g O2 ( 1 mol O2 ) = 1.428125 mol O2

32.00 g O2

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  • 0.2197 mol C4H10 ( 58.14 g C4H10 ) =12.773 → 12.8 g C4H10

1 mol C4H10

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  • 1.0986mol H2O (6.02x 1023 molecules H2O) =

1 mol H2O

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6.6135x1023 → 6.61x1023 molecules H2O

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4 Al +

3 O2 →

2 Al2O3

B

EXCESS

0.78125 mol

0 mol

C

-1.041667 mol

-0.78125 mol

+0.52083 mol

A

EXCESS

0 mol

0.52083 mol

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2 C4H10

13 O2

8 CO2

10 H2O

B

0.2197 mol

1.4281 mol

0 mol

0 mol

C

-0.2197 mol

-1.4281 mol

+0.8789mol

+1.0986mol

A

0 mol

0 mol

0.8789mol

1.0986mol

3 of 12

STATION 3

STATION 4

When steam (hot water) if passed over iron, hydrogen gas and iron (III) oxide are formed. Consider this reaction when 100.0 moles of iron reacts to completion. Steam also reacts to completion.

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___ H2O(g) + ___Fe(s) → ___H2(g) + ___Fe2O3(s)

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  • Create a BCA table

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  • How many grams of steam were needed in this reaction?

10.0 g of solid zinc react with aqueous hydrochloric acid to produce hydrogen gas and aqueous zinc (II) chloride. Assume both reactants react to completion.

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___Zn(s) + ___HCl(aq) → ___H2(g) + ___ZnCl2(aq)

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  • Create a BCA table

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  • What mass of hydrochloric acid was required for this reaction?

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  • Calculate the mass of hydrogen gas produced.

4 of 12

STATION 3 KEY

STATION 4 KEY

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  • Moles is given, so we can plug it directly into out BCA table.

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Did you balance your chemical equation?

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  • 150.0 mol H2O ( 18.02 g H2O ) = 2703 g H2O

1 mol H2O

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  • 10.0 g O2 ( 1 mol Zn ) = 0.15288 mol Zn

65.41 g Zn

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Don’t forget to balance your chemical equation!

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  • 0.30576mol HCl ( 36.46 g HCl ) =11.14800→ 11.1 g HCl

1 mol HCl

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  • 0.15288 mol H2 ( 2.02 g H2 ) = 0.308817 → 0.309 g H2

1 mol H2

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Zn +

2HCl →

H2

ZnCl2

B

0.15288 mol

0.30576 mol

0 mol

0 mol

C

-0.15288mol

-0.30576mol

+0.15288mol

+0.15288mol

A

0 mol

0 mol

0.15288 mol

0.15288 mol

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3 H2O +

2 Fe →

3 H2 +

Fe2O3

B

150.0 mol

100.0 mol

0 mol

0 mol

C

-150.0 mol

-100.0 mol

+150.0 mol

+ 50.00 mol

A

0 mol

0 mol

150.0 mol

50.00 mol

5 of 12

STATION 5

STATION 6

Aqueous ammonium hydroxide reacts with 75.0 g of copper (II) nitrate according to the chemical equation below. Assume both reactants react to completion.

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__NH4OH(aq) + __Cu(NO3)2(aq) → __Cu(OH)2(s) + __NH4NO3(aq)

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  • Create a BCA table

​

  • How many grams of ammonium hydroxide reacted?

1.75x1015 formula units of copper (II) oxide decompose completely to produce solid copper and oxygen gas.

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___CuO(s) → ___Cu(s) + ___O2(g)

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  • Create a BCA table

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  • How many grams of copper metal will be produced?

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  • How many molecules of oxygen gas are produced?

6 of 12

STATION 5 KEY

STATION 6 KEY

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  • 75.0 g Cu(NO3)2 ( 1 mol Cu(NO3)2 ) = 0.400 mol Cu(NO3)2

187.56 g Cu(NO3)2 (don’t round yet!)

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Don’t forget to balance!

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  • 0.800 mol NH4OH (35.04 g NH4OH ) = 28.032 → 28.0 g NH4OH

1 mol NH4OH

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  • 1.75 x 1015 f.u. CuO ( 1 mol CuO ) = 2.90698 x 10-9 mol CuO

6.02 x 1023 f.u. CuO (don’t round yet!)

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Don’t forget to balance!

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  • 2.90698x10-9 mol Cu ( 63.55 g Cu ) = 1.884738x10-7 →

1 mol Cu

1.88x10-7 formula units CuO

  • 1.45349 x 10-9 mol O2 (6.02x 1023 molecules O2) = 8.7500x1014 →

1 mol O2 8.75 x 1014

molecules O2

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2CuO →

2Cu +

O2

B

2.90698x10-9 mol

0 mol

0 mol

C

-2.90698x10-9 mol

+2.90698x10-9 mol

+1.45349 x 10-9 mol

A

0 mol

2.90698x10-9 mol

1.50 x 10-9 mol

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2NH4OH +

Cu(NO3)2 →

Cu(OH)2 +

2NH4NO3

B

0.800 mol

0.400 mol

0 mol

0 mol

C

-0.800 mol

-0.400 mol

+0.400 mol

+ 0.800 mol

A

0 mol

0 mol

0.400 mol

0.800 mol

7 of 12

STATION 7

STATION 8

When ammonia (nitrogen trihydride) is completely burned in the presence of excess oxygen gas, nitrogen gas and water vapor are produced. Suppose 2.32x1022 molecules of nitrogen gas are PRODUCED…

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4 NH3(g) + 3 O2(g) → 2 N2(g) + 6 H2O(g)

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  • Create a BCA table

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  • What mass of ammonia is needed for this reaction to happen?

An acid-base reaction occurs with 25.0 g of sulfuric acid and an unknown amount of sodium hydroxide to produce water and sodium sulfate. Assume both reactants are completely used.

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2 NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + 2 H2O(l)

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  • Create a BCA table

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  • What mass of sodium hydroxide was used in this reaction?

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  • How many grams of sodium sulfate are produced?

8 of 12

STATION 7 KEY

STATION 8 KEY

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  • 2.32x1022 molecules N2 ( 1 mol N2 ) =

6.02x 1023 molecules N2

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0.038538 mol NH3 (don’t round yet!)

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  • 0.077076 mol NH3 ( 17.04 g NH3 ) = 1.313375 → 1.31 g NH3

1 mol NH3

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  • 10.0 g O2 ( 1 mol Zn ) = 0.15288 mol Zn (don’t round yet!)

65.41 g Zn

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Don’t forget to balance your chemical equation!

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  • 0.50973 mol NaOH ( 40.00 g NaOH ) = 20.3896 → 20.4 g NaOH

1 mol NaOH

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  • 0.25487 Na2SO4 ( 119.06 g Na2SO4 ) = 30.3448→ 30.3 g Na2SO4

1 mol Na2SO4

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2NaOH +

H2SO4 →

Na2SO4 +

2 H2O

B

0.50974

0.25487mol

0 mol

0 mol

C

-0.50974 mol

-0.25487 mol

+0.25487 mol

+0.50974 mol

A

0 mol

0 mol

0.25487mol

0.50973mol

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4 NH3 +

3 O2 →

2 N2 +

6 H2O

B

0.077076mol

EXCESS

0 mol

0 mol

C

-0.077076mol

-0.057807 mol

+ 0.038538 mol

+ 0.11561 mol

A

0 mol

EXCESS

0.038538 mol

0.11561 mol

9 of 12

STATION 9

STATION 10

When sodium carbonate reacts completely with excess hydrochloric acid, the carbonic acid that is formed immediately breaks down into carbon dioxide and water. Suppose this reaction PRODUCES 38.5 g of carbon dioxide…

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Na2CO3(s) + 2 HCl(aq) → 2 NaCl(aq) + H2O(l) + CO2(g)

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  • Create a BCA table

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  • How many formula units (particles) of sodium carbonate reacted in this reaction?

3.562 moles of solid sodium metal reacts to completion with water to produce aqueous sodium hydroxide and hydrogen gas.

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___Na(s) + ___H2O(l) → ___NaOH(aq) + ___H2(g)

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  • Create a BCA table

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  • Calculate the molecules of hydrogen gas that are produced.

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  • How many grams of water are needed for the sodium to react completely?

10 of 12

STATION 9 KEY

STATION 10 KEY

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  • 38.5 g CO2 ( 1 mol CO2 ) = 0.8748011 mol CO2

44.01 g CO2 (don’t round yet!)

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  • 0.8748 mol Na2CO3 ( 6.02x1023 form units Na2CO3 ) =

1 mol Na2CO3

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5.26629x1023 → 5.27x1023 formula units of Na2CO3

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  • Moles is given, so you can plug it directly into the BCA table!

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Don’t forget to balance your chemical equation!

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  • 1.781 mol H2 ( 6.02x1023 form units H2 ) =

1 mol H2

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1.07216x1024 → 1.072x1024 molecules H2

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  • 3.562 mol H2O ( 18.02 g H2O ) = 64.18724→ 64.19 g H2O

1 mol H2O

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2 Na +

2 H2O →

2 NaOH +

H2

B

3.562 mol

3.562 mol

0 mol

0 mol

C

- 3.562 mol

- 3.562 mol

+ 3.562 mol

+1.781 mol

A

0 mol

0 mol

3.562 mol

1.781 mol

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Na2CO3 +

2 HCl →

2 NaCl +

H2O +

CO2

B

0.8748 mol

EXCESS

0 mol

0 mol

0 mol

C

-0.8748 mol

-1.7496 mol

+1.7496 mol

+0.8748 mol

+0.8748 mol

A

0 mol

EXCESS

1.7496 mol

0.8748 mol

0.8748 mol

11 of 12

STATION 11

STATION 12

30.0 g of solid iron react with excess copper (I) nitrate solution.

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___CuNO3(aq) + ___Fe(s) → ___Fe(NO3)2(aq) + ___Cu(s)

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  • Create a BCA table

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  • How many grams of copper metal are produced in this reaction?

Ammonia gas (nitrogen trihydride) reacts with 4.5x1022 molecules of nitrogen monoxide (NO). Assume both reactants react to completion.

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4 NH3(g) + 6 NO(g) → 5 N2(g) + 6 H2O(g)

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  • Create a BCA table

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  • Calculate the grams of ammonia (NH3) needed to for this reaction.

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  • What mass of water is produced in this reaction?

12 of 12

STATION 11 KEY

STATION 12 KEY

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  • 30.0 g Fe ( 1 mol Fe ) = 0.537153 mol Fe

55.85 g Fe (don’t round yet!)

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  • 1.074306 mol Cu ( 63.55 g Cu ) = 68.27215 → 68.3 g Cu

1 mol Cu

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  • 4.5x1022 molecules NO ( 1 mol NO ) = 0.0747508 mol NO 6.02x 1023 molecules NO Don’t round yet!

Don’t forget to balance your chemical equation!

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  • 0.049834 mol NH3 ( 17.04 g NH3 ) = 0.84917136 → 0.85 g NH3

1 mol NH3

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  • 0.074751 mol H2O ( 18.02 g H2O ) = 1.34701302 → 1.3 g H2O

1 mol H2O

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4 NH3 +

6 NO →

5N2 +

6 H2O

B

0.049834 mol

0.074751 mol

0 mol

0 mol

C

- 0.049834 mol

- 0.074751 mol

+ 0.062292

+ 0.074751 mol

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A

0 mol

0 mol

0.062292 mol

0.074751 mol

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2 CuNO3 +

Fe →

Fe(NO3)2 +

2 Cu

B

1.074306 mol

0.537153 mol

0 mol

0 mol

C

- 1.074306 mol

- 0.537153 mol

+ 0.537153 mol

+ 1.074306 mol

A

0 mol

EXCESS

0.537153 mol

1.074306 mol