# Solving Problems And Performing a Quantitative Analysis

Industrial Chemistry > 6. The Solvay process has been in use since the 1860s > Solving Problems And Performing a Quantitative Analysis >

Process information to solve problems and quantitatively analyse the relative quantities of reactants and products in each step of the process

PROBLEM 1: Find the masses of limestone and brine needed to produce 1,000 kg of sodium carbonate?

• Step 1: Write the balanced equation.

CaCO3 (s) + 2 NaCl (aq) à Na2CO3 (s) + CaCl2 (aq)

• Step 2: Find the number of moles of sodium carbonate.

Using the equation, n = m/M, where:
n= number of moles (mol)
m= mass (g)
M= molar mass (g/mol)

n = m/M
n = 1,000,000g / 105.99 g/mol Na2CO3
n = 9,435 mol

Looking at the mole to mole ratio, NaCl: Na2CO3 = 2:1. Therefore, there are 18,870 moles of NaCl needed to produce 1,000 kg of Na2CO3.

• Step 3: Find the mass of the starting material.

Manipulating the equation, n = m/M,

m= n × M
m = 18,870 mol NaCl × 58.44 g/mol NaCl
m= 1,103 kg of NaCl

Using the mole to mole ratio again, CaCO3: Na2CO3 shows 1:1 ratio, then,

m= n × M
m = 9,435 mol CaCO3× 100.09 g/mol CaCO3
m= 944 kg of CaCO3

SOLUTION: Therefore, 1,103 kg and 944 kg of NaCl and CaCO3, respectively, were needed to produce 1,000 kg of Na2CO3.

PROBLEM 2: During the preparation of calcium oxide, 1,000 kg of limestone (calcium carbonate) is heated. What mass of calcium oxide is produced and calculate the volume of carbon dioxide produced if the molar volume = 24.5 L/mol?

• Step 1: Write the balanced equation.

CaCO3 (s) à CaO (s) + CO2 (g)

HEAT

• Step 2: Find the number of moles of calcium carbonate.

Using the equation, n = m/M, where:

n= number of moles (mol)
m= mass (g)
M= molar mass (g/mol)

n = m/M
n = 1,000,000g / 100.09 g/mol CaCO3
n = 9,991 mol CaCO3

Looking at the mole to mole ratio, CaCO3: CaO = 1:1. Therefore, there is also 9,991 moles of CaO that will be produced for every 1,000 kg of CaCO3.

• Step 3: Find the mass of the starting material.

Manipulating, the equation n = m/M,

m= n × M
m = 9,991 mol CaO × 56.08 g/mol CaO
m= 560 kg of CaO

Using the mole to mole ratio again, CaCO3: CO2 shows 1:1 ratio.  Therefore, there is also 9,991 moles of CO2 that will be produced for every 1,000 kg of CaCO3. By using the formula, n= V/Vm, where:

n= number of moles (mol)
V= volume (L)
Vm= molar volume (L/mol)

V = n × Vm
V = 9,991 mol CO2 × 24.5 L/ mol
V = 245,780 L

SOLUTION: Therefore, 560 kg and 245,789 L of CaO and CO2, respectively, were produced from 1,000 kg of CaCO3.