how many grams of al are required to generate 16.35 l of h₂ gas when the reaction occurs at stp?\n2al(s) +…

how many grams of al are required to generate 16.35 l of h₂ gas when the reaction occurs at stp?\n2al(s) + 6hcl(aq) → 2alcl₃(aq) + 3h₂(g)\nmass of al = ? g\ndo not round until the end.

how many grams of al are required to generate 16.35 l of h₂ gas when the reaction occurs at stp?\n2al(s) + 6hcl(aq) → 2alcl₃(aq) + 3h₂(g)\nmass of al = ? g\ndo not round until the end.

Answer

Explanation:

Step1: Moles of H₂ at STP

At STP, 1 mole of gas = 22.4 L. $$n_{H_2} = \frac{16.35\ \text{L}}{22.4\ \text{L/mol}}$$

Step2: Mole ratio Al:H₂

From reaction, 2 mol Al → 3 mol H₂. $$n_{Al} = n_{H_2} \times \frac{2\ \text{mol Al}}{3\ \text{mol H}_2}$$

Step3: Mass of Al (molar mass=26.98 g/mol)

$$m_{Al} = n_{Al} \times 26.98\ \text{g/mol}$$ Substitute $n_{H_2}$: $$m_{Al} = \frac{16.35}{22.4} \times \frac{2}{3} \times 26.98$$

Answer:

12.9 g