Chemistry
posted by Hannah .
What volume of o.410 M HNO3 would you need to neutralize 18.75mL of 0.376M Ba(OH)2?
Would you do (0.410)(18.75) / 0.376?
Or (0.410)(0.376) / 18.75

Neither.
(0.376)(18.75)(2)/(0.410)= ?
Here is how you do it.
Write and balance the equation.
2HNO3 + Ba(OH)2 ==> 2H2O + Ba(NO3)2
moles Ba(OH)2 initially = M x L = 0.00705 mols (or 18.75 x 0.376 = 7.05 millimoles).
Now use the coefficients in the balanced equation to convert moles Ba(OH)2 to moles HNO3. That will be
moles Ba(OH)2 x (2 moles HNO3/1 mole Ba(OH)2) = 7.05 mmoles x 2 = 14.10 millimoles HNO3.
Now convert to L or mL HNO3.
M = moles/L or M = millimoles/mL.
0.410 = 14.10/mL Solve for mL.
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