Answer
178 g Na
Work Step by Step
Balanced chemical reaction:
2Na(s) +2 $H_{2}O$ (l)${\longrightarrow}$ 2 NaOH (aq) +$H_{2}$ (g)
The strategy to find the mass of Na required to produce 7.81 g $H_{2}$ is:
g $H_{2}$ ${\rightarrow}$ mol $H_{2}$ ${\rightarrow}$ mol Na ${\rightarrow}$ gNa
The conversion factors are:
$\frac{1 mol (H_{2}}{Molar mass (H_{2})}$ to convert to moles $H_{2}$, where Molar mass $H_{2}$:
Molar mass $H_{2}$= 2 x 1.00 = 2.00 g
$\frac{2 mol (Na)}{1 mol (H_{2})}$, to convert the moles of $H_{2}$ to mol of Na.
$\frac{Molar mass(Na)}{1 mol(Na)}$ to convert to g of Na, where the molar mass of Na is:
molar mass of Na = 1 x 22.90 = 22.9 g
Now we use of the conversion factors to find the mass of Na:
7.81 g $H_{2}$ x $\frac{1 mol (H_{2}}{Molar mass (H_{2})}$ x $\frac{2 mol (Na)}{1 mol (H_{2})}$ x $\frac{Molar mass(Na)}{1 mol(Na)}$ =
7.81 g $H_{2}$ x $\frac{1 mol (H_{2}}{2.0g (H_{2})}$ x $\frac{2 mol (Na)}{1 mol (H_{2})}$ x $\frac{22.9(Na)}{1 mol(Na)}$ = 178 g Na
