Question

Calculate the maximum mass of sodium carbonate that could be formed by reacting 40 kg of calcium carbonate with an excess of sodium chloride solution.

(Relative formula masses: CaCO3 = 100; Na2CO3 = 106)

Answer 1

Answer:

2NaCl + CaCO3 → Na2CO3+ CaCl2

Explanation:

Answer 2

The maximum mass of sodium carbonate that is formed by reacting 40 kg of calcium carbonate with an excess of sodium chloride solution is 42.4 Kg.

Given,

Mass of calcium carbonate that reacts=40 Kg

Mass of CaCO3=100 g

Mass of Na2CO3=106 g.

To find,

the maximum mass of sodium carbonate that could be formed by reacting 40 kg of calcium carbonate with an excess of sodium chloride solution.

Solution:

• The balanced chemical reaction between calcium carbonate and sodium chloride can be given as:
• 2NaCl + CaCO3 → Na2CO3+ CaCl2.
• The stochiometric coefficients of NaCl, CaCO3 and Na2CO3 are 2, 1 and 1 respectively.
• This suggests that 2 moles of NaCl react with 1 mole of CaCO3 to give 1 mole of Na2CO3.

Moles of calcium carbonate reaction will be:

$Moles=\frac{mass}{molar mass} \\Moles=\frac{40X1000}{100} \\Moles=400.$

If  1 mole of CaCO3 gives 1 mole of Na2CO3, then 400 moles of CaCO3 will give 400 moles of Na2CO3.

Mass of Na2CO3 obtained will be:

$Moles=\frac{mass}{molar mass} \\Maa=MolesXMolar mass\\Mass=400X106 g\\Mass=42400 g\\Mass=\frac{42400}{1000} \\Mass=42.4 Kg.$

Hence, the mass of Na2CO3 obtained is 42.4 Kg.

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