Based on the number of moles of H_2SO_4, determine the number of moles of Na_2CO_3 needed for the neutralization reaction.
Sodium carbonate ( Na_2CO_3) is used to neutralize the sulfuric acid spill. How many kilograms of sodium carbonate must be added to neutralize 4.03×103 kg of sulfuric acid solution?
3.90 * 10^4 = H_2SO_4 mol
Mole to mole ratio is 1:1.
To determine the number of moles of Na2CO3 needed for the neutralization reaction, we can use the mole-to-mole ratio between H2SO4 and Na2CO3, which is 1:1.
Given that the number of moles of H2SO4 is 3.90 * 10^4 moles, we can conclude that the number of moles of Na2CO3 needed is also 3.90 * 10^4 moles.
However, in the second part of your question, you are asked to determine the mass of Na2CO3 needed to neutralize a certain mass of sulfuric acid.
To solve this, we need to use the molar mass of Na2CO3, which is 105.99 g/mol.
First, let's convert the mass of sulfuric acid into moles. Given that the molar mass of sulfuric acid (H2SO4) is 98.09 g/mol, we can determine the number of moles:
molar mass H2SO4 = 98.09 g/mol
mass H2SO4 = 4.03 * 10^3 kg = 4.03 * 10^6 g
moles H2SO4 = mass H2SO4 / molar mass H2SO4
moles H2SO4 = (4.03 * 10^6 g) / (98.09 g/mol)
Now, using the mole-to-mole ratio of 1:1 between H2SO4 and Na2CO3, we can determine the mass of Na2CO3 needed:
moles Na2CO3 = moles H2SO4
mass Na2CO3 = moles Na2CO3 * molar mass Na2CO3
mass Na2CO3 = (4.03 * 10^6 g / 98.09 g/mol) * 105.99 g/mol
Therefore, the mass of Na2CO3 needed to neutralize 4.03 * 10^3 kg of sulfuric acid is (4.03 * 10^6 g / 98.09 g/mol) * 105.99 g/mol.