what mass of sodium sulphate is produced if 49g of hydrogen sulphate react with 80g of NAOH?
2 answers
is this from CA
2NaOH + H2SO4 --> Na2SO4 + 2H2O
This is a limiting reagent (LR) problem. You know that when an amount is given for more than one reactant. I work these the long way.
mols NaOH = g/molar mass = 80/40 = 2
2 mols NaOH will produce 1 mol Na2SO4 from the equation so we get 1 mol Na2SO4.
mols H2SO4 = 49/98 = 0.5
From the equation, 1 mol H2SO4 will produce 1 mol Na2SO4 so we get 0.5 mol Na2SO4 from this reagent.
Both numbers can't be right. In LR problems we always get the smaller number; therefore, we will produce 0.5 mols Na2SO4.
Then grams Na2SO4 = mols Na2SO4 x molar mass Na2SO4 = ?
Post your work if you get stuck.
This is a limiting reagent (LR) problem. You know that when an amount is given for more than one reactant. I work these the long way.
mols NaOH = g/molar mass = 80/40 = 2
2 mols NaOH will produce 1 mol Na2SO4 from the equation so we get 1 mol Na2SO4.
mols H2SO4 = 49/98 = 0.5
From the equation, 1 mol H2SO4 will produce 1 mol Na2SO4 so we get 0.5 mol Na2SO4 from this reagent.
Both numbers can't be right. In LR problems we always get the smaller number; therefore, we will produce 0.5 mols Na2SO4.
Then grams Na2SO4 = mols Na2SO4 x molar mass Na2SO4 = ?
Post your work if you get stuck.
1 answer