Question

Consider the reaction: 2 NaBH4(aq) + H2SO4(aq) → 2 H2(g) + Na2SO4(aq) + B2H6(g) What volume, in mL, of a 0.526 M solution of NaBH4 is required to produce 0.574 g of B2H6? H2SO4 is present in excess.

Answer 1

no of moles of B2H6   = W/G.M.Wt

                                   = 0.574/27.66   = 0.02075moles

2 NaBH4(aq) + H2SO4(aq) → 2 H2(g) + Na2SO4(aq) + B2H6(g)

1 mole of B2H6 produced from 2 moles of NaBH4

0.02075 moles of produced from = 2*0.02075/1   = 0.0415 moles of NaBH4

no of moles of NaBH4 = molarity *volume in L

0.0415                          =0.526*volume in L

volume in L                  = 0.0415/0.526   = 0.0789L

volume of solution = 78.9ml >>>>answer