__Solutions
to____ Stoichiometry__

#### 1. For the reaction: 2K + 0.5 O_{2} à 1K_{2}O

a. How many moles of O_{2} are
needed to react with 0.56 moles of K?

0.56 moles of K

b. How many moles of O_{2} are needed
to make 7.6 g of K_{2}O?

7.6 g K_{2}O (mole/94g K_{2}O)
= 0.0809 moles K_{2}O

0.0809 moles K_{2}O

c. How many grams of K_{2}O will
be produced from 0.50 g of K?

0.50
g K (mole/39g K) = 0.0128 moles K_{}

0.0128 moles K

x = 0.0064 moles K_{2}O (78+16)g/mole
= 0.60 g

#### 2. For the reaction: Na_{2}O + H_{2}O à2 NaOH

a. What mass NaOH could be made from 12.4
g of Na_{2}O?

12.4 g of Na_{2}O (mole/62 g) = 0.2
mole Na_{2}O

0.2 mole Na_{2}O

0.4 moles NaOH(40 g/mole) = 16 g NaOH

b. How many moles of Na_{2}O are needed
to make 1000 g of NaOH?

1000g (mole/40 g) = 25 moles NaOH

25 moles NaOH

c. What would happen if 18 g of water were
mixed with 18 g of sodium oxide?

18 g of H_{2}O is 1 mole

18 g of Na_{2}O = 18g(mole/62g) =0.29 moles

but according to the equation water and sodium oxide react in a 1:1, so we
have too much water . All 0.29 moles of Na_{2}O will react with only
0.29 moles of water to create twice as many moles of NaOH (see ratio). Answer =
0.58 moles of NaOH result.

0.58 moles(40
g/mole) = 23.2 g

#### 3. Balance
and answer the questions that follow:

#### C + 2 H_{2}àCH_{4}

a. How many moles of CH_{4} can be
made from 7.0 g of H_{2}?

7.0 g of H_{2}= 3.5 moles H_{2. }How? Divide by molar mass of diatomic hydrogen._{}

From the ratio, only half as many moles of
methane will be produced so answer = 1.75 moles CH_{4}.

b. What weight of H_{2} is needed
to react with 5.0 g of C?

Convert 5.0 g of C to moles: 0.417 moles C

From ratio: 0.834 moles of H_{2} is needed.

Convert to mass by multiplying by molar
mass:

1.66 g H_{2}

c.
What would happen if 20 g of hydrogen were mixed with 20 g of carbon?

20
g of C(mole/12 g) = 1.67 mole C. This amount needs
2(1.75 moles) = 3 moles of H_{2}.(see ratio).
We clearly have an excess of H_{2} since 20 g = 10 moles.

The amount of CH_{4} produced will depend on C, the
limiting reagent, and only 1.67e moles of CH_{4 }will be produced.

1.67 moles (16g/mole) = 27 g