Question

How many grams of sodium oxide can be produced when 55.3 g Na react with 64.3 g O2?

Answer 1

Answer : The amount of sodium oxide produced is, 74.40 grams

Solution : Given,

Mass of sodium = 55.3 g

Mass of $O_2$= 64.3 g

Molar mass of sodium = 23 g/mole

Molar mass of $O_2$ = 32 g/mole

Molar  mass of $Na_2O$ = 61.9 g/mole

First we have to calculate the moles of sodium and oxygen.

Moles of Na = $\frac{\text{ given mass of Na}}{\text{ molar mass of Na}}= \frac{55.3g}{23g/mole}=2.404moles$

Moles of $O_2$ = $\frac{\text{ given mass of }O_2}{\text{ molar mass of }O_2}= \frac{64.3g}{32g/mole}=2.009moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction will be,

$4Na+O_2\rightarrow 2Na_2O$

From the balanced reaction we conclude that

4 moles of sodium react with 1 mole of oxygen

2.404 moles of sodium react with $\frac{2.404}{4}=0.601$ moles of oxygen

Excess moles of oxygen = 2.009 - 0.601 = 1.408 moles

That means sodium is a limiting reagent and oxygen is an excess reagent.

Now we have to calculate the moles of sodium oxide.

From the reaction we conclude that,

4 moles of sodium react to give 2 moles of sodium oxide

2.404 moles of sodium react to give $\frac{2}{4}\times 2.404=1.202$ moles of sodium oxide

Now we have to calculate the mass of sodium oxide.

$\text{Mass of }Na_2O=\text{Moles of }Na_2O\times \text{Molar mass of }Na_2O$

$\text{Mass of }Na_2O=(1.202moles)\times (61.9g/mole)=74.40g$

Therefore, the amount of sodium oxide produced is, 74.40 grams

Answer 2

Given the two amounts of reactants, we get the limiting reactant between the two. We convert each to mole and divide them to their corresponding stoichiometric coefficient. we balance first above, that is 4Na+O2=2Na2O. hence, for Na that is 0.60 mole and oxygen is 2.00 mole. We follow the smaller amount, Na, hence the amount of sodium oxide in grams is 74.53 grams.