2Al + 3H2SO4 ==> Al2(SO4)3 + 3H2
mols Al = g/atomic mass
mols H2SO4 = M x L. L = 0.075 and M = 1.18 x 1000 x 0.247 x (1/98) = ?
From the problem I suppose we assume H2SO4 is the excess reagent. Use stoichiometry to determine H2SO4 used, subtract from the initial amount to find that remaining, the M of the remaining -= mols/L solution.
Post your work if you get stuck.
A piece of aluminium weighing 2.7 gm is heated with 75 ml of H2SO4(specific gravity 1.18 containing 24.7 percent H2SO4
by weight.)After the metal is carefully dissolved,the solution is dissolved to 400ml.Calculate tne molarity of the free H2SO4 in the resulting solution.(IIT 1992)
2 answers
A 25-mL sample of 0.160M solution of NaOH is titrated with 17 mL of an unknown solution of H2SO4. What is the molarity of the sulfuric acid solution? A. 0.004M H2SO4 B. 0.235M H2SO4 C. 0.117M H2SO4 D. 0.002M H2SO4
2 answers