Asked by cinthia sabillon
How many mL of 0.4040 M HNO3 would be needed to neutralize 55.00 mL of 0.1234 M potassium hydroxide?
Answers
Answered by
DrBob222
KOH + HNO3 ==> KNO3 + H2O
mols KOH = M x L = ?
mols HNO3 = mols KOH (you know from the coefficients in the balanced equation; that is 1:1)
M HNO3 = mols HNO3/L HNO3. You know M and mols, solve for L and convert to mL.
mols KOH = M x L = ?
mols HNO3 = mols KOH (you know from the coefficients in the balanced equation; that is 1:1)
M HNO3 = mols HNO3/L HNO3. You know M and mols, solve for L and convert to mL.
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