Q. 7.59

Question

Sodium reacts with oxygen to produce sodium oxide.

a. How many grams of $N a_{2} O$ are produced when $57.5 g$ of $N a$ reacts?

b. If you have $18.0 g$ of $N a$ , how many grams of $O_{2}$ are needed for the reaction?

c. How many grams of $O_{2}$ are needed in a reaction that produces $75.0 g$ of $N a_{2} O$ ?

Step-by-Step Solution

Verified

Answer

a. $77.5 g$ of $N a_{2} O$ produced from the reaction.

b. $6.24 g$ of $O_{2}$ is required for the reaction.

c. $19.35 g$ or $O_{2}$ is required for the reaction.

1Part (a) Step 1: Given Information

We need to find the grams of $N a_{2} O$ which produced when $57.5 g$ of $N a$ reacts.

2Part (a) Step 2: Calculation

The mole factor is given as

$m o l e f a c t o r = \frac{2 m o l e N a_{2} O}{4 m o l e N a}$

We know that

mass of $N a = 57.5 g$ ,

molar mass of $N a = 23 g / m o l e$

so,

$m o l e o f N a = \frac{m a s s}{m o l a r m a s s} = \frac{57.5 g}{23 g / m o l e} = 2.5 m o l e$

We have

$4 m o l e o f N a = 2 m o l e o f N a_{2} O (2.5 m o l e N a) \times (\frac{2 m o l e N a_{2} O}{4 m o l e N a}) = 1.25 m o l e N a_{2} O$

Now,

$M a s s o f N a_{2} O = (m o l a r m a s s) \times (m o l e) = 62 \frac{g}{m o l e} \times 1.25 m o l e M a s s o f N a_{2} O = 77.5 g$

3Part (b) Step 1: Given Information

We need to find how many grams of $O_{2}$ required to react with $18.0 g$ of $N a$ .

4Part (b) Step 2: Calculation

The mole factor is given as

$m o l e f a c t o r = \frac{1 m o l e O_{2}}{4 m o l e N a}$

We have

mass of $N a = 18 g$ ,

molar mass of $N a = 23 g / m o l e$

$m o l e o f N a = \frac{m a s s}{m o l a r m a s s} = \frac{18 g}{23 g / m o l e} = 0.78 m o l e$

$4 m o l e N a = 1 m o l e O_{2} (0.78 m o l e N a) \times (\frac{1 m o l e O_{2}}{4 m o l e N a}) = 0.195 m o l e O_{2}$

Now,

$m a s s o f O_{2} = (m o l a r m a s s) \times (m o l e) = 32 \frac{g}{m o l e} \times 0.195 m o l e m a s s o f O_{2} = 6.24 g$

5Part (c) Step 1: Given Information

We need to find the grams of $O_{2}$ require for the reaction to produce $75.0 g$ of $N a_{2} O$ .

6Part (c) Step 2: Calculation

The mole factor is given as

$m o l e f a c t o r = \frac{1 m o l e O_{2}}{2 m o l e N a_{2} O}$

We have

mass of $N a_{2} O = 75.0 g$ ,

molar mass of $N a_{2} O = 62 g / m o l e$

So,

$m o l e o f N a_{2} O = \frac{m a s s}{m o l a r m a s s} = \frac{75 g}{62 g / m o l e} m o l e o f N a_{2} O = 1.21 m o l e$

also we have

$2 m o l e N a_{2} O = 1 m o l e O_{2}$

so,

$(1.21 m o l e N a_{2} O) \times (\frac{1 m o l e O_{2}}{2 m o l e N a_{2} O}) = 0.604 m o l e O_{2}$

Now,

$m a s s o f O_{2} = (m o l a r m a s s) \times (m o l e) = 32 \frac{g}{m o l e} \times 0.604 m o l e m a s s o f O_{2} = 19.35 g$

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