<span>5.45 x 10^3 kg of sodium carbonate is needed to neutralize 5.04 kg of sulfuric acid.
For this, I will assume you have pure H2SO4. So first, you need to calculate the molar mass of H2SO4 and Na2CO3. Lookup the atomic weights of all the elements involved.
Atomic weight of Sodium = 22.989769
Atomic weight of Sulfur = 32.065
Atomic weight of Carbon = 12.0107
Atomic weight of Oxygen = 15.999
Atomic weight of Hydrogen = 1.00794
Molar mass of H2SO4 = 2 * 1.00794 + 32.065 + 4 * 15.999 = 98.07688 g/mol
Molar mass of Na2CO3 = 2 * 22.989769 + 12.0107 + 3 * 15.999
= 105.987238 g/mol
The balanced equation for the reaction of Na2CO3 with H2SO4 is
Na2CO3 + H2SO4 => Na2SO4 + CO2 + H2O
so for every mole of sulfuric acid to be neutralized, you need 1 mole of sodium carbonate. You can determine the number of moles of sulfuric acid you have and then calculate the mass of that many moles of sodium carbonate. But, there's an easier way. Just use the relative mass differences between sodium carbonate and sulfuric acid. So
105.987238 g/mol / 98.07688 g/mol = 1.080655
So that means for every kg of sulfuric acid, you need 1.080655 kg of sodium carbonate. Now do the multiplication.
5.04 x 10^3 kg * 1.080655 = 5.4465 x 10^3 kg.
Since you only have 3 significant figures for your data, round the result to 3 significant figures, giving 5.45 x 10^3 kg</span>
It was Erwin Schrodinger.
Answer:
The answer to your question is:
Explanation:
Ideal gases
PV = nRT
P = nRT / V
P = (1)(0.082)(303) / 0.5
P = 49.7 atm
Real pressure
P = V²(nRT)/ (V - nb) - an²
P = (0.5)²(1)(0.082)(303) / (0.5 - (1)(0.03219) - (1.345)(1)²
P = (6.2115)/(0.4678) - 1.345
P = 13.28 - 1.345
P = 11.93 atm
