Question

consider the reaction: 2 nabh4(aq) + h2so4(aq) → 2 h2(g) + na2so4(aq) + b2h6(g). what volume, in ml, of a 0.496 m solution of nabh4 is required to produce 0.515 g of b2h6? h2so4 is present in excess.

Explanation:

Step1: Calculate moles of B₂H₆

The molar mass of B₂H₆ is \(M_{B_2H_6}=(2\times10.81 + 6\times1.01)\text{ g/mol}=27.68\text{ g/mol}\). The number of moles of B₂H₆, \(n_{B_2H_6}=\frac{m}{M}\), where \(m = 0.515\text{ g}\). So \(n_{B_2H_6}=\frac{0.515\text{ g}}{27.68\text{ g/mol}}=0.0186\text{ mol}\)

Step2: Determine moles of NaBH₄

From the balanced - chemical equation \(2NaBH_4(aq)+H_2SO_4(aq)
ightarrow2H_2(g)+Na_2SO_4(aq)+B_2H_6(g)\), the mole ratio of \(NaBH_4\) to \(B_2H_6\) is \(n_{NaBH_4}:n_{B_2H_6}=2:1\). So the number of moles of \(NaBH_4\) required, \(n_{NaBH_4}=2\times n_{B_2H_6}=2\times0.0186\text{ mol}=0.0372\text{ mol}\)

Step3: Calculate volume of NaBH₄ solution

The molarity of the \(NaBH_4\) solution is \(M = 0.496\text{ M}\). Using the formula \(M=\frac{n}{V}\) (where \(V\) is volume in liters), we can solve for \(V\). Rearranging the formula gives \(V=\frac{n}{M}\). Substituting \(n = 0.0372\text{ mol}\) and \(M = 0.496\text{ M}\), we get \(V=\frac{0.0372\text{ mol}}{0.496\text{ mol/L}} = 0.075\text{ L}\)

Step4: Convert volume to mL

Since \(1\text{ L}=1000\text{ mL}\), then \(V = 0.075\text{ L}\times1000\text{ mL/L}=75\text{ mL}\)

Answer:

75