Question

1. Either write the balanced equation or balance the given equation. Then, solve the problem.

1.
_____NaCl +______ H₂SO4 →
HCI + Na₂SO4
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a) What is the mass, in grams, of sodium chloride that reacts with 275.0 g of sulfuric acid?
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b) If 12.3 mol HCl are produced in this reaction, how many grams of sodium sulfate are produced?

Answer 1

2 NaCl + H₂SO4 → 2 HCl + Na₂SO4

a) 327.8 g of NaCl.

b)874.8 g Na2SO4

Step-by-step explanation:

To balance this equation,

2NaCl + H2S04 -> 2 HCl + Na2SO4.

a)To find the mass of sodium chloride, we need to determine the limiting reactant.

275.0g H2SO4 x (1mol of H2SO4/ 98.08g of H2SO4) = 2.804 mol of H2SO4.

So we need, 2x 2.804 = 5.608 moles of NaCl to react with Sulphuric Acid. Then convert it to grams using the molar mass of NaCl,

5.608 moles of NaCl x 58.44g of NaCl/mol = 327.8 g NaCl.

b) To find how many grams of sodium sulfate are produced, we can

12.3 mol HCl × (1 mol Na2SO4 / 2 mol HCl) = 6.15 mol Na2SO4

To find the mass of Na2SO4 produced, we can use its molar mass:

6.15 mol Na2SO4 × 142.04 g Na2SO4/mol = 874.8 g Na2SO4

Therefore, 874.8 g of sodium sulfate is produced when 12.3 mol HCl is produced in this reaction.