Question

How many grams of hydrogen gas can be produced from the following reaction when $65$ grams of zinc and $65$ grams of HCl are allowed to react.
$Zn\left(s\right)+2HCl\left(aq\right)\to ZnC{l}_2\left(aq\right)+H_2\left(g\right)$

• A. $1.0$
• B. $1.8$
• C. $3.6$
• D. $7.0$
• E. $58$

Answer 1

The correct option is B $1.8$

$Zn\left(s\right)+2HCl\left(aq\right)\to ZnC{l}_2\left(aq\right)+H_2\left(g\right)$
1 mole of Zn reacts with 2 moles of HCl to produce 1 mole of hydrogen.
The atomic mass of Zn is 65.4 g/mol. The molar mass of HCl is 36.5 g/mol.
When we divide mass with atomic/molar mass, we get number of moles.
Number of moles of Zn $=\frac{65}{65.4}=0.994$ moles.
Number of moles of HCl $=\frac{65}{36.5}=1.78$ moles.
0.994 moles of Zn will react with $2\times 0.994=1.988$ moles of HCl however only 1.78 moles of HCl are present. Hence, HCl is the limiting reactant.
1.78 moles of HCl will produce $\frac{1.78}{2}=0.89$ moles of $H_2$.
The molar mass of $H_2$ is 2 g/mol.
Mass of $H_2$ obtained $=2g/mol\times 0.89mol=1.78g=1.8g$