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How many mL of 0.376M barium hydroxide are needed to completely neutralize 94.72mL of 0.374M nitric acid?HELP!

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How many mL of 0.376M barium hydroxide are needed to completely neutralize 94.72mL of 0.374M nitric acid?

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  1. Moles HNO3 = 0.09472 L x 0.374 =0.0354

    2 HNO3 + Ba(OH)2 >> Ba(NO3)2 + 2 H2O

    moles Ba(OH)2 = 0.0354/2 =0.0177

    V = 0.0177/0.376 =0.0471 L =>47.1 mL

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