How many mL of 10% aq. sodium carbonate are needed to neutralize 1.5 mL of sulfu

Question

How many mL of 10% aq. sodium carbonate are needed to neutralize 1.5 mL of sulfuric acid? (Sulfuric acid is 18M.) I already know the answer is 28 ml of sodium carbonate. (18 x 1.5) / .94 = 20 mL I have no idea where the .94 came from, will someone please show me the math on this. How many mL of 10% aq. sodium carbonate are needed to neutralize 1.5 mL of sulfuric acid? (Sulfuric acid is 18M.) I already know the answer is 28 ml of sodium carbonate. (18 x 1.5) / .94 = 20 mL I have no idea where the .94 came from, will someone please show me the math on this. How many mL of 10% aq. sodium carbonate are needed to neutralize 1.5 mL of sulfuric acid? (Sulfuric acid is 18M.) I already know the answer is 28 ml of sodium carbonate. (18 x 1.5) / .94 = 20 mL I have no idea where the .94 came from, will someone please show me the math on this.

Explanation / Answer

H2SO4 + Na2CO3 -----------> Na2SO4 + CO2 + H2O

moles of H2SO4 = 18 x 1.5/1000 = 0.027  

mass of H2SO4 = 0.027 x 98 = 2.646 g

according to balanced reaction

98 g H2SO4 reacts with 106 g Na2CO3

2.646 g H2SO4 reacts with 2.646 x 106 / 98 = 2.862 g

mass of Na2CO3 require = 2.862 g

% m/v = (mass of solute / volume of solution) x 100

10 = (2.862 / volume of solution) x 100

0.1 = (2.862 / volume of solution)

volume of solution = 28.62 mL

volume of Na2CO3 required = 28.62 mL

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