Question

1. H₂ is produced by the reaction of 119.1 mL of a 0.9147 M solution of H₃PO₄ according to the following equation.

     2 Cr + 2 H₃PO₄ → 3 H₂ + 2 CrPO₄

(a) Determine the number of moles of H₂.   ------mol

(b) Determine the mass (in g) of H₂. -------g

2. Molecules of I₂ are produced by the reaction of 0.4245 mol of CuCl₂ according to the following equation.

      2 CuCl₂ + 4 KI → 2 CuI + 4 KCl + I₂

(a) How many molecules of I₂ are produced?

(b)What mass (in g) of I₂ is produced? ------g

Answer 1

Calculate moles of H3PO4

Mol = volume in L x molarity

Volume is 0.1191 L and molarity is 0.9147 M

Mol = 0.1191*0.9147=0.1089 mol H3PO4

Determine mol ratio between H3PO4 and H2

2 mol H3PO4 : 3 mol H2

Moles of H2 formed = moles of H3PO4 x 3 mol H2 / 2 mol H3PO4

0.1089*3/2=0.1634 mol H2

Mass of H2

Mass = moles x molar mass

Molar mass of H2 is 2.0158 g/mol

0.1634*2.0158=0.3294 g

Mass of H2 produced is 0.3294 g

2).

a) determine mol ratio between CuCl2 : I2

Its 2 : 1

Calculate moles using moles of CuCl2

Mol I2 = 0.4245 mol CuCl2 x 1 mol I2 / 2 mol CuCl2

0.4245*1/2=0.2123 mol I2

b) mass of I2

Molar mass of I2 = 253.8 g/mol

Mass of I2 = 0.2123 x 253.8=53.88 g

So the mass of I2 produced would be 53.88 g