# In the reaction between sodium oxide and water, what mass of sodium oxide will be needed to produce 5.90g of sodium hydroxide ?

Apr 29, 2018

$5.25 \text{g}$

#### Explanation:

First we need our balanced equation.

$N {a}_{2} O + {H}_{2} O \rightarrow 2 N a O H$

For this problem, we need to work backwards, finding moles of $2 N a O H$, then how many moles of $N {a}_{2} O$ was required to make the $2 N a O H$, then convert that to grams.

First we find how many moles of $2 N a O H$ are in 5.90g.

$5.90 \text{g" * "1mol"/"40.00g" = .148"g}$

Then we check how many moles of $N {a}_{2} O$ were required to produce it.

$0.148 \text{mol "Na_2O" * "1mol Na_2O"/"2"mol 2NaOH" = 0.0740"mol}$

Then convert the moles to grams.

$\text{0.0740mol" * 70.99"g/mol" = 5.25"g}$