Answer
1.46 x x $10^{5}$ g W
Work Step by Step
Balanced chemical reaction:
$WO_{3}$(s) + 3 $H_{2}$(g) ${\longrightarrow}$ W(s) + 3 $H_{2}O$(g)
The strategy to find the mass of W that can be obtained from 4.81 kg $H_{2}$ is:
g $H_{2}$ ${\rightarrow}$ mol $H_{2}$ ${\rightarrow}$ mol W ${\rightarrow}$ g W
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 1.01 g = 2.02 g
$\frac{1 mol (W)}{3 mol (H_{2})}$, to convert the moles of $H_{2}$ to mol of W.
$\frac{Molar mass(W)}{1 mol(W)}$ to convert to g of W, where the molar mass of W is 183.8 g
Now we use of the conversion factors to find the mass of W:
4.81 x $10^{3}$ g $H_{2}$ x $\frac{1 mol (H_{2})}{Molar mass (H_{2})}$ x $\frac{1 mol (W)}{3 mol (H_{2})}$ x $\frac{Molar mass(W)}{1 mol(W)}$ =
4.81 x $10^{3}$ g $H_{2}$ x $\frac{1 mol (H_{2})}{2.02 g (H_{2})}$ x $\frac{1 mol (W)}{3 mol (H_{2})}$ x $\frac{183.8 g(W)}{1 mol(W)}$ = 146 x $10^{3}$ g W or 1.46 x x $10^{5}$ g W
