Stoichiometry Worksheet Answers

1.

a. 2 / 13

b. 13 / 8

c. 13 / 10

d. 2 / 8 (or 1 / 4)

e. 2 / 10 (or 1 / 5)

2. The KClO_{3} / O_{2} molar ratio is 2/3.

2 mol KClO_{3} / 3 mol. O_{2} = 12.00 mol KClO_{3} / x = 18.00 mol.

x = 18.00 mol of O_{2}

3. Given the following equation: 2 K + Cl_{2} ---> 2 KCl

How many grams of KCl is produced from 2.50 g of K and excess Cl_{2}.

From 1.00 g of Cl_{2} and excess K?

4. Given the following equation: Na_{2}O + H_{2}O ---> 2 NaOH

How many grams of NaOH is produced from 1.20 x 10^{2} grams of Na_{2}O?

How many grams of Na_{2}O are required to produce 1.60 x 10^{2} grams of NaOH?

5. Given the following equation: 8 Fe + S_{8} ---> 8 FeS

What mass of iron is needed to react with 16.0 grams of sulfur?

How many grams of FeS are produced?

6. Given the following equation: 2 NaClO_{3} ---> 2 NaCl + 3 O_{2}

12.00 moles of NaClO_{3} will produce how many grams of O_{2}?

How many grams of NaCl are produced when 80.0 grams of O_{2} are produced?

7. Given the following equation: Cu + 2 AgNO_{3} ---> Cu(NO_{3})_{2} + 2 Ag

How many moles of Cu are needed to react with 3.50 moles of AgNO_{3}?

If 89.5 grams of Ag were produced, how many grams of Cu reacted?

8. Molten iron and carbon monoxide are produced in a blast furnace by the reaction of iron(III) oxide and coke (pure carbon). If 25.0 kilograms of pure Fe_{2}O_{3} is used, how many kilograms of iron can be produced? The reaction is: Fe_{2}O_{3} + 3 C ---> 2 Fe + 3 CO

9. The average human requires 120.0 grams of glucose (C_{6}H_{12}O_{6}) per day. How many grams of CO_{2} (in the photosynthesis reaction) are required for this amount of glucose? The photosynthetic reaction is: 6 CO_{2} + 6 H_{2}O ---> C_{6}H_{12}O_{6} + 6 O_{2}

This problem is slightly different from those above.

10. Given the reaction: 4 NH_{3} (g) + 5 O_{2} (g) ---> 4 NO (g) + 6 H_{2}O (l)

When 1.20 mole of ammonia reacts, the total number of moles of products formed is:

a. 1.20 b. 1.50 c. 1.80 d. 3.00 e. 12.0

The correct answer is d.

NH_{3} / (NO + H_{2}O) = 4 / 10

4 / 10 = 1.20 / x

x = 3.00 mol