Question

For each of the following balanced chemical equations, calculate how many moles of product(s) would be produced if 0.640 mol of the first reactant were to react completely.

(a) BaCl₂(aq) + 2 AgNO₃(aq) → 2 AgCl(s) + Ba(NO₃)₂(aq)

(b) S(s) + 2 H₂SO₄(aq) → 3 SO₂(g) + 2 H₂O(l)

(c) C₃H₈(g) + 5 O₂(g) → 4 H₂O(l) + 3 CO₂(g)

(d) BaCl₂(aq) + H₂SO₄(aq) → BaSO₄(s) + 2 HCl(aq)

Answer 1

(a) BaCl₂(aq) + 2 AgNO₃(aq) → 2 AgCl(s) + Ba(NO₃)₂(aq)

Moles of BaCl₂ = 0.640

Moles of AgCl = 0.640 moles BaCl2 * 2 mole of AgCl / 1 mole BaCl2

= 1.28 moles AgCl

Moles of Ba(NO₃)₂= 0.640 moles BaCl2 * 1 mole of Ba(NO₃)₂/ 1 mole BaCl2

= 0.640 moles Ba(NO₃)₂

(b) S(s) + 2 H₂SO₄(aq) → 3 SO₂(g) + 2 H₂O(l)

Moles of S= 0.640 moles

Moles of SO2 = 0.640moles S * 3 mole SO2/ 1 mole S

= 1.92 moles SO2

Moles of H₂O = 0.640moles S * 2 mole H₂O / 1 mole S

= 1.28 moles H₂O

(c) C₃H₈(g) + 5 O₂(g) → 4 H₂O(l) + 3 CO₂(g)

Moles of C3H8 = 0.640 moles

Moles of H₂O = 0.640moles C3H8 * 4 mole H₂O / 1 mole C3H8

= 2.56   moles H₂O

Moles of CO₂ = 0.640moles C3H8 * 3 mole CO₂ / 1 mole C3H8

= 1.92 moles CO₂

(d) BaCl₂(aq) + H₂SO₄(aq) → BaSO₄(s) + 2 HCl(aq)

Moles of BaCl₂ = 0.640

Moles of BaSO₄ = 0.640 moles BaCl2 * 1 mole of BaSO₄ / 1 mole BaCl2

= 0.640   moles BaSO₄

Moles of HCl = 0.640 moles BaCl2 * 2 mole of HCl / 1 mole BaCl2

= 1.28 moles HCl