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.

CaCO_{3 (s)} + 2 NaCl _(aq) à Na₂CO_{3 (s)} + CaCl_{2 (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.

CaCO_{3 (s)} à CaO _(s) + CO_{2 (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/V_m, where:

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

V = n × V_m
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.