Question

Sodium borohydride is normally produced by the reaction of sodium hydride with trimethyl borate according to the following equation.

NaH + B(OCH3)3 ® NaBH4 + NaOCH3

a) Balance the above equation. Be sure to get this correct since your answer to part (b) depends on your correct balancing of the equation.

b) If 18.3 g of NaH are reacted with 18.5 g of B(OCH3)3, what is the maximum mass of NaBH4 that can be produced?

Answer 1

a) Balanced equation

4NaH + B(OCH3)3 -------> NaBH4 + 3 NaOCH3

b) Number of moles = Mass in gram / Molar mass

Mass of NaH = 18.3 g, Molar mass of NaH = 24 g/mol

Number of moles of NaH = 18.3 g / 24 g/mol = 0.7625 mol

Mass of B(OCH3)3 = 18.5 g, Molar mass = 103.91 g/mol

Number of moles of B(OCH3)3 = 18.5 g / 103.91g/mol = 0.178 mol

From reaction, 1.0 mol of B(OCH3)3 reacts with 4.0 mol of NaH produces 1.0 mol of NaBH4 then 0.178 mol of B(OCH3)3 will react with 4 * 0.178 mol = 0.712 mol of NaH will produce 0.178 mol of NaBH4.

Molar mass of NaBH4 = 37.83 g/mol

Mass of NaBH4 can produced = 37.83 g/mol * 0.178 mol = 6.73 g