Final answer:
Approximately 6.54×103 kg of sodium carbonate is needed to neutralize 6.05×103 kg of sulfuric acid, assuming a 1:1 mole ratio from the balanced equation. This answer is not reflected in the given options, suggesting there may be an error in the question.
Step-by-step explanation:
The student has asked how many kilograms of sodium carbonate (Na2CO3) would be required to neutralize a sulfuric acid spill of 6.05×103 kg. To answer this, we need to consider the balanced chemical equation for the reaction between sulfuric acid (H2SO4) and sodium carbonate (Na2CO3), which is:
Na2CO3 + H2SO4 → Na2SO4 + CO2 + H2O
From the equation, we can see that one mole of sodium carbonate reacts with one mole of sulfuric acid to neutralize it. The molar mass of Na2CO3 is approximately 106 g/mol, while that of H2SO4 is approximately 98 g/mol.
To convert the mass of sulfuric acid to moles, we use the formula:
mass (in kg) × (1000 g/kg) / molar mass (in g/mol) = moles
For 6.05×103 kg of H2SO4:
6.05×103 kg × (1000 g/kg) / 98 g/mol = 6.17×104 moles
This means we need 6.17×104 moles of Na2CO3 to neutralize the sulfuric acid. Converting this to kilograms:
moles × molar mass (in g/mol) / (1000 g/kg) = kg
6.17×104 moles × 106 g/mol / (1000 g/kg) ≈ 6.54×103 kg
Therefore, to neutralize 6.05×103 kg of sulfuric acid, approximately 6.54×103 kg of sodium carbonate would be required. This is not one of the options provided, possibly indicating a typo or error in the question or its options. A professional should recheck the calculations and options given in the question.