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AleksandrR [38]
3 years ago
11

A tanker truck carrying 6.05×103 kg of concentrated sulfuric acid solution tips over and spills its load. The sulfuric acid solu

tion 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
Chemistry
1 answer:
irinina [24]3 years ago
3 0

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.

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