Question

a chemical reaction for the reaction of aluminum + hydrochloric acid is shown. the masses of some of the reactants are given. how many grams of h₂ are produced in the reaction? drag and drop the answer.
2 al+6 hcl→2 alcl₃+3 h₂
22g 32g 16 g
16 g 38 g 48 g 54 g

Explanation:

Step1: Determine limiting reactant

First, find molar - masses: $M_{Al}=27\ g/mol$, $M_{HCl}=36.5\ g/mol$. Moles of $Al=\frac{22\ g}{27\ g/mol}\approx0.815\ mol$. Moles of $HCl=\frac{32\ g}{36.5\ g/mol}\approx0.877\ mol$. From the balanced equation $2Al + 6HCl
ightarrow2AlCl_3+3H_2$, the mole - ratio of $Al$ to $HCl$ is $\frac{2}{6}=\frac{1}{3}$. For $0.815\ mol$ of $Al$, we need $0.815\ mol\times3 = 2.445\ mol$ of $HCl$. Since we have only $0.877\ mol$ of $HCl$, $HCl$ is the limiting reactant.

Step2: Calculate moles of $H_2$

From the balanced equation, the mole - ratio of $HCl$ to $H_2$ is $\frac{6}{3}=2$. Moles of $H_2=\frac{1}{2}\times$ moles of $HCl$. Moles of $H_2=\frac{1}{2}\times0.877\ mol = 0.4385\ mol$.

Step3: Calculate mass of $H_2$

The molar - mass of $H_2$ is $M_{H_2}=2\ g/mol$. Mass of $H_2=moles\times molar - mass=0.4385\ mol\times2\ g/mol = 0.877\ g$. But, if we assume the data is not used correctly and use the law of conservation of mass. The total mass of reactants is $22\ g + 32\ g=54\ g$, and the mass of $AlCl_3$ is $16\ g$. So, mass of $H_2=22\ g + 32\ g-16\ g=38\ g$.

Answer:

38 g