What mass of Na2so4 is produced if 49g of h2so4 react with 80g of Naoh
2Naoh + h2so4 Na2so4 + 2h2o
To determine the mass of Na2SO4 produced, we need to first determine the limiting reactant. The limiting reactant is the reactant that is completely consumed and determines the maximum amount of product that can be formed.
1. Calculate the number of moles of each reactant:
- Moles of H2SO4 = mass / molar mass = 49g / (1 mol H2SO4/98.09g) = 0.499 mol H2SO4
- Moles of NaOH = mass / molar mass = 80g / (1 mol NaOH/39.997g) = 2 mol NaOH
2. Determine the stoichiometric ratio between H2SO4 and Na2SO4:
From the balanced chemical equation, we can see that the stoichiometric ratio between H2SO4 and Na2SO4 is 1:1.
3. Calculate the maximum moles of Na2SO4 that can be produced:
Based on the stoichiometric ratio, the maximum moles of Na2SO4 that can be produced is also 0.499 mol.
4. Determine the mass of Na2SO4 produced:
Mass of Na2SO4 = moles * molar mass = 0.499 mol * (1 mol Na2SO4/142.04g) = 0.352g
Therefore, 0.352g of Na2SO4 is produced.