Calculate the volume, in mL, of a 1.22 M Na₂S solution needed to react with 2.75 g of AlBr₃ in the following equation:
3 Na₂S (aq) + 2 AlBr₃ (aq) $\rightarrow$ Al₂S₃ (s) + 6 NaBr (aq)

Question

Calculate the volume, in mL, of a 1.22 M Na₂S solution needed to react with 2.75 g of AlBr₃ in the following equation: 3 Na₂S (aq) + 2 AlBr₃ (aq)$\rightarrow$Al₂S₃ (s) + 6 NaBr (aq)

Quizplus · Accepted Answer

First, calculate moles of AlBr3: Molar mass of AlBr3 = 266.69 g/mol, so moles = 2.75 g / 266.69 g/mol = 0.01031 mol. From the balanced equation, 2 moles of AlBr3 react with 3 moles of Na2S, so moles of Na2S needed = 0.01031 mol * (3/2) = 0.015465 mol. Volume of Na2S = moles / concentration = 0.015465 mol / 1.22 M = 0.0127 L = 12.7 mL.

Calculate the Volume, in ML, of a 1 →\rightarrow→ Al2S3 (S) + 6 NaBr (Aq) A) 0

Calculate the Volume, in ML, of a 1 $\rightarrow$ Al₂S₃ (S) + 6 NaBr (Aq)
A) 0