How many milliliters of 1.58 M HCl are needed to react completely with 23.2 g of NaHCO₃(ℳ = 84.02 g/mol)? HCl(aq) + NaHCO₃(s) → NaCl(s) + H₂O(l) + CO₂(g) A)638 mL B)572 mL C)536 mL D)276 mL E)175 mL

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Quizplus · Accepted Answer

First, calculate moles of NaHCO₃: 23.2 g / 84.02 g/mol = 0.276 mol. The reaction is 1:1, so 0.276 mol of HCl is needed. Using the concentration of HCl (1.58 M), calculate volume: 0.276 mol / 1.58 mol/L = 0.175 L, which is 175 mL.

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