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The image shows a chemistry question asking for the volume of 0.02 m aqueous HBr required to neutralize 10.0 ml of 0.01 m aqueous Ba(OH)2, assuming complete neutralization. The options provided are:rn(a) 2.5 mlrn(b) 7.5 mlrn(c) 5.0 mlrn(d) 10. ml
Accepted Answer
Balanced chemical equation:
2 HBr(aq) + Ba(OH)2(aq) → 2 H2O(l) + BaBr2(aq)
Moles of Ba(OH)2 = (0.01 mol/L) * (10.0 mL/1000 mL/L) = 1.00 x 10^-4 mol
Moles of HBr required = 2 * moles of Ba(OH)2 = 2.00 x 10^-4 mol
Volume of 0.02 M HBr required = (2.00 x 10^-4 mol) / (0.02 mol/L) = 10.0 mL
Therefore, the correct answer is (D) 10. mL
2 HBr(aq) + Ba(OH)2(aq) → 2 H2O(l) + BaBr2(aq)
Moles of Ba(OH)2 = (0.01 mol/L) * (10.0 mL/1000 mL/L) = 1.00 x 10^-4 mol
Moles of HBr required = 2 * moles of Ba(OH)2 = 2.00 x 10^-4 mol
Volume of 0.02 M HBr required = (2.00 x 10^-4 mol) / (0.02 mol/L) = 10.0 mL
Therefore, the correct answer is (D) 10. mL
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