H2SO4 (aq) + 2 NaOH(aq) ==> Na2SO4(aq) + 2 H2O (I)
This is a limiting reagent (LR) problem. You know that when quantities are given for BOTH reactants. I'm estimating the numbers so you should redo all of them.
mols H2SO4 = g/molar mass = 4/98 = about 0.04
mols NaOH = 4/40 = 0.10
If H2SO4 and excess NaOH were used then Na2SO4 produced would be 0.04 x (1 mol Na2SO4/1 mol H2SO4) = 0.04 mols Na2SO4.
If NaOH and excess H2SO4 were used then Na2SO4 produced would be
0.10 x (1 mol Na2SO4/2 mols NaOH) = 0.05 mols Na2SO4
In LR problems the smaller number always is the correct value so 0.04 mols Na2SO4 will be produced. Convert to grams = mols Na2SO4 x molar mass Na2SO4. Remember to recalculate all of the numbers.
When solutions of H2SO4 and NaOH react, the balanced molecular equation is: H2SO4 (aq) + 2 NaOH(aq) Na2SO4(aq) + 2 H2O (I)
How much Na2SO4 is obtained when 4.00 g of H2SO4 reacts with 4.00 g of NaOH?
2 answers
BTW, I don't see much organic chemistry in this problem.
2 answers