Question

A tanker truck carrying 6.05×103 kg of concentrated sulfuric acid solution tips over and spills its load. The sulfuric acid solution is 95.0%H2SO4 by mass and has a density of 1.84 g/mL.
Part A

Sodium carbonate (Na2CO3) is used to neutralize the sulfuric acid spill. How many kilograms of sodium carbonate must be added to neutralize 6.05×103 kg of sulfuric acid solution?

Express your answer with the appropriate units

Answer 1

Answer:

6,216.684 kilograms of sodium carbonate must be added to neutralize $6.05\times 10^3 kg$ of sulfuric acid solution.

Explanation:

Mass of sulfuric acid solution = $6.05\times 10^3 kg=6.05\times 10^6 g$

$1 kg = 10^3 g$

Percentage mass of sulfuric acid = 95.0%

Mass of sulfuric acid = $\frac{95.0}{100}\times 6.05\times 10^6 g$

$=5,747,500 g$

Moles of sulfuric acid = $\frac{5,747,500 g}{98 g/mol}=58,647.96 mol$

$H_2SO_4+Na_2CO_3\rightarrow Na_2SO_4+CO_2+H_2O$

According to reaction , 1 mole of sulfuric acid is neutralized by 1 mole of sodium carbonate.

Then 58,647.96 moles of sulfuric acisd will be neutralized by :

$\frac{1}{1}\times 58,647.96 mol=58,647.96 mol$ of sodium carbonate

Mass of 58,647.96 moles of sodium carbonate :

$106 g/mol\times 58,647.96 mol=6,216,683.76 g$

6,216,683.76 g = 6,216,683.76 × 0.001 kg = 6,216.684 kg

6,216.684 kilograms of sodium carbonate must be added to neutralize $6.05\times 10^3 kg$ of sulfuric acid solution.