Asked by Charles
What volume of 3.00 mol/L HNO3 is needed to neutralize 450 mL of 0.100 mol/L Sr(OH)2
Answers
Answered by
DrBob222
2HNO3 + Sr(OH)2 ==> Sr(NO3)2 + 2H2O
mols Sr(OH)2 = M x L - 0.100 M x 0.450 = 0.0450
From the equation you see it will take 2 mol HNO3 for every 1 mol Sr(OH)2 so it will take 0.0450 x 2 = 0.0900 mol HNO3.
Then M HNO3 = mols/L . You know mols and you know M, solve for L. Convert to mL if you wish.
mols Sr(OH)2 = M x L - 0.100 M x 0.450 = 0.0450
From the equation you see it will take 2 mol HNO3 for every 1 mol Sr(OH)2 so it will take 0.0450 x 2 = 0.0900 mol HNO3.
Then M HNO3 = mols/L . You know mols and you know M, solve for L. Convert to mL if you wish.
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