Calculate the mass of lithium hydroxide that must be added to 50ml of water so that it may react exactly with 5ml solution of 1.5M nitric acid.
2 answers
2.416g
I don't get that. You should show your work so we can show you where you are making an error.
LiOH + HNO3 ==> LiNO3 + H2O
mols HNO3 = M x L = 1.5M x 0.005L = 0.0075 mols.
mols LiOH = 0.0075
g LiOH = mols x molar mass = 0.0075 x 23.95 = 0.1796 g LiOH which rounds to 0.18 g LiOH to two significant figures or to 0.2g to one s.f. I can't tell from your problem how many s.f. you have. The 5 mL limits the problem to 1 but you may have omitted the zero from 5.0 mL. The 1.5 limits the problem to 2.
LiOH + HNO3 ==> LiNO3 + H2O
mols HNO3 = M x L = 1.5M x 0.005L = 0.0075 mols.
mols LiOH = 0.0075
g LiOH = mols x molar mass = 0.0075 x 23.95 = 0.1796 g LiOH which rounds to 0.18 g LiOH to two significant figures or to 0.2g to one s.f. I can't tell from your problem how many s.f. you have. The 5 mL limits the problem to 1 but you may have omitted the zero from 5.0 mL. The 1.5 limits the problem to 2.