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12.5 mL * 5.0 (m)mol/(m)L HCl = 62.5 mmol spilled HCl

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62.5 mmol NaHCO3 = 62.5 mmol * 84.01 (m)g/(m)mol NaHCO3 = 5250 mg NaHCO3 = 5.25 g pure NaHCO3

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Q: What mass of NaHCO3 would be needed to neutralize a spill consisting of 12.5 mL of 5.0 M HCl?
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