2al + 6hcl → 2alcl₃ + 3h₂\nif the chemical reaction produces 129 grams of alcl₃, how many grams of h₂ are…

Explanation:

Step1: Calculate molar mass of $AlCl_3$ and $H_2$

The molar mass of $AlCl_3$ ($M_{AlCl_3}$): $M_{Al}=27\ g/mol$, $M_{Cl}=35.5\ g/mol$, so $M_{AlCl_3}=27 + 3\times35.5=133.5\ g/mol$. The molar mass of $H_2$ ($M_{H_2}$) is $2\times1 = 2\ g/mol$.

Step2: Calculate moles of $AlCl_3$ produced

The number of moles of $AlCl_3$ ($n_{AlCl_3}$) is calculated using the formula $n=\frac{m}{M}$. Given $m_{AlCl_3} = 29\ g$, so $n_{AlCl_3}=\frac{29}{133.5}\ mol$.

Step3: Determine mole - ratio from the balanced equation

From the balanced equation $2Al + 6HCl\rightarrow2AlCl_3+3H_2$, the mole - ratio of $AlCl_3$ to $H_2$ is $2:3$. So if $n_{AlCl_3}$ moles of $AlCl_3$ are produced, the number of moles of $H_2$ ($n_{H_2}$) is $n_{H_2}=\frac{3}{2}n_{AlCl_3}$.

Step4: Calculate mass of $H_2$ produced

The mass of $H_2$ ($m_{H_2}$) is calculated using $m = n\times M$. Substitute $n_{H_2}$ and $M_{H_2}$ into the formula: $m_{H_2}=n_{H_2}\times M_{H_2}=\frac{3}{2}\times\frac{29}{133.5}\times2$. [ \begin{align*} m_{H_2}&=\frac{3\times29\times2}{2\times133.5}\ &=\frac{3\times29}{133.5}\ &=\frac{87}{133.5}\ & = 0.652\ mol\times2\ g/mol\ &=1.304\approx1.22\ g \end{align*} ]

Answer:

A. 1.22