The following reaction shows calcium chloride reacting with silver nitrate

CaCl2 + AgNO3 - >

AgCl + Ca (NO3) 2

How many grams of AgCl are produced from 30.0 grams of CaCl2?

Ca = 40.078, Cl = 35.453,

O = 15.999, Ag = 107.868,

N = 14.007

A. 19.0

B. 38.8

C. 58.2

D. 77.5

D. 77.5.

Explanation:

From the balanced equation: CaCl₂ + 2AgNO₃ → 2AgCl + Ca (NO₃) ₂ It is clear that 1.0 mole of CaCl₂ reacts with 2.0 moles of AgNO₃ to produce 2.0 moles of AgCl and 1.0 mole of Ca (NO₃) ₂. The no. of moles of reacted (30.0 g) CaCl₂ = mass / molar mass = (30.0 g) / (110.98 g/mol) = 0.27 mol.

Using cross multiplication:

1.0 mole of CaCl₂ produces → 2.0 moles of AgCl, from the stichiometry.

0.27 mol of CaCl₂ produces →? moles of AgCl.

∴ The no. of moles of the produced AgCl = (2.0 mol) (0.27 mol) / (1.0 mol) = 0.54 mol.

∴ The mass of the produced AgCl = no. of moles x molar mass = (0.54 mol) (143.32 g/mol) = 77.48 g ≅ 77.5 g.