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C1 JUN 22 Q9d

The Solvay process is the major industrial process for the manufacture of sodium
carbonate. Two of the stages in the process are shown below:
NaHCO3+ NH4CI
NaCl + NH3 + COz + H,O
2NaHCO3
-
-
Na₂CO3 + H₂O + CO₂
I
Calculate the maximum mass of sodium carbonate, in kg, which could be obtained
from 15.0 tonnes of sodium chloride. Give your answer to an appropriate number of
significant figures.
[3]

User Skycrew
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1 Answer

3 votes

Answer:

The balanced chemical equation for the reaction between sodium chloride and ammonia bicarbonate is:

NaHCO3 + NH4Cl → NaCl + NH3 + CO2 + H2O

From the equation, we see that one mole of sodium chloride reacts with one mole of ammonia bicarbonate to produce one mole of sodium bicarbonate and one mole of ammonium chloride. The balanced equation for the decomposition of sodium bicarbonate to sodium carbonate is:

2NaHCO3 → Na2CO3 + H2O + CO2

From the equation, we see that two moles of sodium bicarbonate decompose to produce one mole of sodium carbonate.

We start by calculating the number of moles of sodium chloride in 15.0 tonnes:

Number of moles of NaCl = mass ÷ molar mass

Number of moles of NaCl = 15000 kg ÷ 58.44 g/mol

Number of moles of NaCl = 2.568 × 10^5 mol

From the balanced equation, we know that the same number of moles of sodium bicarbonate and ammonium chloride are produced. Therefore, the number of moles of sodium bicarbonate produced is also 2.568 × 10^5 mol.

Now, we need to convert the number of moles of sodium bicarbonate to the maximum mass of sodium carbonate that could be produced. Since two moles of sodium bicarbonate produce one mole of sodium carbonate, we can use the molar mass of sodium carbonate to calculate the mass:

Mass of Na2CO3 = number of moles of NaHCO3 ÷ 2 × molar mass of Na2CO3

Mass of Na2CO3 = (2.568 × 10^5 mol ÷ 2) × 105.99 g/mol

Mass of Na2CO3 = 1.358 × 10^7 g

Mass of Na2CO3 = 1.358 × 10^4 kg

Therefore, the maximum mass of sodium carbonate that could be obtained from 15.0 tonnes of sodium chloride is 1.358 × 10^4 kg, or 1.36 × 10^4 kg to three significant figures.

User Jon Rubins
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8.7k points