what mass of h2 forms when 35.25 g of al reacts with excess hydrochloric acid? 2al + 6hcl → 2alcl3 + 3h2 al…
what mass of h2 forms when 35.25 g of al reacts with excess hydrochloric acid? 2al + 6hcl → 2alcl3 + 3h2 al : 26.98 g/mol h2 : 2.02 g/mol ? g h2
Answer
Answer:
3.94 g
Explanation:
Step1: Calculate moles of Al
$n_{Al}=\frac{m_{Al}}{M_{Al}}=\frac{35.25\ g}{26.98\ g/mol}\approx1.307\ mol$
Step2: Determine mole - ratio
From the balanced equation $2Al + 6HCl\rightarrow2AlCl_{3}+3H_{2}$, the mole - ratio of $Al$ to $H_{2}$ is $\frac{n_{H_{2}}}{n_{Al}}=\frac{3}{2}$.
Step3: Calculate moles of $H_{2}$
$n_{H_{2}}=\frac{3}{2}n_{Al}=\frac{3}{2}\times1.307\ mol = 1.9605\ mol$
Step4: Calculate mass of $H_{2}$
$m_{H_{2}}=n_{H_{2}}\times M_{H_{2}}=1.9605\ mol\times2.02\ g/mol\approx3.94\ g$