Diborane ( B 2 H 6) can be prepared by the following reaction- 3 NaBH 4+4 BF 3 → 3 NaBF 4+2 B 2 H 6 If the reaction has a 70 % yield, how many moles of NaBH 4 should be used with excess BF 3 in order to obtain 0.200 mol of B 2 H 6 ?

Question

Diborane ( B 2 H 6) can be prepared by the following reaction- 3 NaBH 4+4 BF 3 → 3 NaBF 4+2 B 2 H 6 If the reaction has a 70 % yield, how many moles of NaBH 4 should be used with excess BF 3 in order to obtain 0.200 mol of B 2 H 6 ? (A) 0.21 moles (B) 0.429 moles (C) 0.300 mol (D) 0.175 moles

Tardigrade Admin · Accepted Answer

3 NaBH 4+4 BF 3 → 3 NaBF 4+2 B 2 H 6 Since BF 3 is in excess, the limiting reagent is NaBH 4. To obtain 2 mole B 2 H 6, NaBH 4 required =3 mole To obtain 0.200 mole B2 H6, NaBH 4 required =(3/2) × 0.200 =0.300 mole Because, the yield is 70 %, Hence x × (70/100)=0.300 x=(0.300/1000) × (100/70)=(3/7)=0.429 moles

Q. Diborane (B2​H6​) can be prepared by the following reaction- 3NaBH4​+4BF3​→3NaBF4​+2B2​H6​ If the reaction has a 70% yield, how many moles of NaBH4​ should be used with excess BF3​ in order to obtain 0.200mol of B2​H6​ ?

0.21 moles

0.429 moles

0.300 mol

0.175 moles

Q. Diborane $(B_{2} H_{6})$ can be prepared by the following reaction-
$3 N a B H_{4} + 4 B F_{3} \to 3 N a B F_{4} + 2 B_{2} H_{6}$
If the reaction has a $70%$ yield, how many moles of $N a B H_{4}$ should be used with excess $B F_{3}$ in order to obtain $0.200 m o l$ of $B_{2} H_{6}$ ?