Question

a) A 4.951 g sample of a metal reacted with sulfuric acid to produce a gas and a sulfate solid. What is the expected mass (in grams) of the dried sulfate if the metal was Al?

b) A 7.77 g sample of a metal reacted with sulfuric acid to produce a gas and a sulfate solid. What is the expected mass (in grams) of the dried sulfate if the metal was Zn?

Answer 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 1: write the balanced chemical equation.

(a) 2Al + 3H₂SO₄ ---> Al₂(SO₄)₃ + 3H₂

(b)   Zn + H₂SO₄ ---> ZnSO₄ + H₂

Step 2: calculate the moles of Al and Zn

we know, moles = mass given / molar mass

(a) moles of Al = ( 4.951 g / 26.981538 g/mol ) = 0.1835 mol

(a) moles of Zn  = ( 7.77 g / 65.38 g/mol ) = 0.119 mol

Step 3: Calculate the moles of sulfate solid

(a) 2Al + 3H₂SO₄ ---> Al₂(SO₄)₃ + 3H₂

According to the reaction:
2 mol of Al produce 1 mol of  Al₂(SO₄)₃
so, 0.1835  mol of Al will produce=( 1 mol of Al₂(SO₄)₃  / 2 mol of Al ) × 0.1835 mol of Al = 0.09175 mol of Al₂(SO₄)₃

(b)   Zn + H₂SO₄ ---> ZnSO₄ + H₂

According to the reaction:
1 mol of Zn produce 1 mol of   ZnSO₄
so, 0.119  mol of Al will produce=( 1 mol of ZnSO₄   / 1 mol of Zn ) × 0.119 mol of Zn = 0.119 mol of ZnSO₄

Step 4: Calculation of dried sulfate produced

(a) we get moles of  Al₂(SO₄)₃  that can be produced = 0.09175  mol

(b) we get moles of  ZnSO₄   that can be produced = 0.119 mol

(a) Mass of Al₂(SO₄)₃  produced =( moles×molar mass ) = ( 0.09175 mol × 342.15 g/mol ) = 31.4 g

(a) Mass of ZnSO₄ produced =( moles×molar mass ) = ( 0.119 mol × 161.47 g/mol ) = 19.2 g