How many grams of Cu2S could be produced from 9.90 g of CuCl with an excess of H2S gas?

The grams of CuS produced is 7.95 mass

calculation

Step 1: write the equation for reaction

2CUCl + H2S → CU2S + 2 HCl

Step 2 : calculate the moles of CUCl

moles = mass/molar mass

The molar mass of CuCl = 63.5 for CU + 35.5 for Cl = 99 g/mol

=9.90 g / 99 g/mol = 0.1 moles

Step 3: use the mole ratio to determine the moles of CU2S

CUCl:CU2S is 2:1 therefore the moles of CU2S = 0.1 x1/2=0.05 moles

step 4: find the mass of CU2S

mass = molar mass x moles

The molar mass of CU2S = (2 x 63.5) + 32 = 159 g/mol

mass = 0.05 x 159 = 7.95 mass

Given:

Mass of CuCl = 9.90 g

To determine:

Mass of Cu2S produced

Explanation:

Balanced reaction-

2CuCl + H2S → Cu2S + 2HCl

In the presence of excess H2S, CuCl will be the limiting reagent

Thus, based on stoichiometry: 2 moles of CuCl will produce 1 mole of Cu2S

Now,

given mass of CuCl = 9.9 g

molar mass of CuCl = 99 g/mol

# moles of CuCl = 9.9/99 = 0.1 moles

Moles of Cu2S produced = 0.1 * 1/2 = 0.05 moles

Molar mass of Cu2S = 159 g/mol

Mass of Cu2S produced = 7.95 g

Ans: Thus the yield of Cu2S is 7.95 g