Question

2.) Calculate the mass (in grams) of H2 produced from the reaction of 34.23 g of Mg with excess HCI. Mg(s) + HCI (aq) → MgCl2 (aq) + H2 (g)

How do I solve this problem step by step?

Answer 1

Answer:

Step 1: Explanation

A mole ratio is ​the ratio between the amounts in moles of any two compounds involved in a chemical reaction. The mole ratio can be determined by examining the coefficients in front of formulas in a balanced chemical equation.

Step 2:  write the balanced chemical equation.

Mg(s) + 2 HCl(aq) ------> MgCl2(s) + H2(g)

Step 3: calculate the moles of Mg

We know,

Molar mass of Mg =24.305 g/mol

Mass given = 34.23 g

Moles of Mg = mass given / molar mass = ( 34.23 g / 24.305 g/mol ) = 1.41 mol

Step 4: Calculate the moles of H2 gas produced

Mg(s) + 2 HCl(aq) ------> MgCl2(s) + H2(g)

According to the reaction:
1 mol of Mg produced from 1 mol of  H2

So, 1.41 mol of Mg will produced from = 1.41 mol of H2

Hence , Moles of H2 produced = 1.41 mol

Step 5: Calculation of mass of H2 gas produced

we get moles of H2 that can be produced = 1.41 mol

Molar mass of H2 = 2.016 g/mol

Mass of H2 produced =( moles × molar mass ) = (1.41 mol × 2.016 g/mol ) = 2.84 g

Hence, the mass of H2 gas produced = 2.84 g