What mass of Na2so4 is produced if 49g of h2so4 react with 80g of Naoh
Naoh + h2so4 Na2so4 + 2h2o
To determine the mass of Na2SO4 produced, we need to find the limiting reactant first. The limiting reactant is the reactant that is completely consumed in a reaction and determines the maximum amount of product that can be formed.
We can use the molar masses and stoichiometry to find the number of moles of each reactant.
1. Calculate the number of moles of H2SO4:
moles of H2SO4 = mass of H2SO4 / molar mass of H2SO4
moles of H2SO4 = 49g / (1 mol H2SO4/98.09g H2SO4)
moles of H2SO4 = 0.5 mol
2. Calculate the number of moles of NaOH:
moles of NaOH = mass of NaOH / molar mass of NaOH
moles of NaOH = 80g / (1 mol NaOH/ 39.997g NaOH)
moles of NaOH = 2 mol
3. Use the stoichiometry of the balanced equation to determine the moles of Na2SO4 produced:
According to the balanced equation, the ratio of H2SO4 to Na2SO4 is 1:1.
moles of Na2SO4 = 0.5 mol H2SO4
Since the stoichiometric ratio of NaOH to Na2SO4 is 2:1, we can multiply the moles of NaOH by 0.5 to find the moles of Na2SO4:
moles of Na2SO4 = (2 mol NaOH / 1 mol Na2SO4) * 0.5 mol NaOH
moles of Na2SO4 = 1 mol Na2SO4
4. Calculate the mass of Na2SO4:
mass of Na2SO4 = moles of Na2SO4 * molar mass of Na2SO4
mass of Na2SO4 = 1 mol * (1 mol Na2SO4/ 142.04g Na2SO4)
mass of Na2SO4 = 142.04g
Therefore, the mass of Na2SO4 produced is 142.04g.