Question

1. hcl reacts with mg to produce mgcl₂ and h₂ gas. if i have 56.2g of hcl, how many moles of mgcl₂ will be produced?

2 hcl + mg → mgcl₂ + h₂

1. how many moles of oxygen are needed to react with 7.5 moles of aluminium?

4 al + 3 o₂ → 2 al₂o₃

Explanation:

(Problem 9):

Step1: Calculate moles of HCl

Molar mass of HCl = $1.008 + 35.45 = 36.458\ \text{g/mol}$
$\text{Moles of HCl} = \frac{56.2\ \text{g}}{36.458\ \text{g/mol}} \approx 1.542\ \text{mol}$

Step2: Mole ratio from balanced equation

From $\ce{2HCl + Mg -> MgCl2 + H2}$, mole ratio $\ce{HCl:MgCl2} = 2:1$
$\text{Moles of } \ce{MgCl2} = \frac{1}{2} \times \text{Moles of HCl}$
$\text{Moles of } \ce{MgCl2} = \frac{1}{2} \times 1.542\ \text{mol} = 0.771\ \text{mol}$

(Problem 10):

Step1: Mole ratio from balanced equation

From $\ce{4Al + 3O2 -> 2Al2O3}$, mole ratio $\ce{Al:O2} = 4:3$
$\text{Moles of } \ce{O2} = \frac{3}{4} \times \text{Moles of Al}$

Step2: Substitute given moles of Al

$\text{Moles of } \ce{O2} = \frac{3}{4} \times 7.5\ \text{mol} = 5.625\ \text{mol}$

Answer:

1. Moles of $\ce{MgCl2}$ produced: $0.771\ \text{mol}$
2. Moles of $\ce{O2}$ needed: $5.625\ \text{mol}$