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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...Asked by Jeconiah
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.
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Answered by
Jeconiah
2.416g
Answered by
DrBob222
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.
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