# How would you calculate the number of grams of sodium oxide, Na2O, that will be produced when 5.00 moles react with oxygen given the reaction: 4Na + O2--> 2 Na2O?

Nov 1, 2015

$4 N a + {O}_{2} \rightarrow 2 N {a}_{2} O$.

By the stoichiometry of this reaction if 5 mol natrium react, then 2.5 mol $N {a}_{2} O$ should result.

#### Explanation:

The molecular mass of natrium oxide is $61.98$ $g \cdot m o {l}^{- 1}$. If $5$ $m o l$ natrium react, then $\frac{5}{2}$ $m o l \times 61.98$ $g \cdot m o {l}^{- 1}$ $=$ $154.95$ $g$ natrium oxide should result.

So what have I done here? First, I had a balanced chemical equation (this is the important step; is it balanced?). Then I used the stoichiometry to get the molar quantity of product, and converted this molar quantity to mass. If this is not clear, I am willing to have another go.