Question

Be sure to answer all parts. Calculate the mass of each product formed when 76.99 g of diborane (B2H) reacts with excess water: B2H6(g) + H20() → H3BO3(s) + H2(g) unbalanced I Mass of H3BO3: Mass of H2:

Answer 1

Balance reaction

B2H6 + H2O = H3BO4 + H2

balanc eB

B2H6 + H2O = 2H3BO4 + H2

balance O

B2H6 + 4H2O = 2H3BO4 + H2

balance H

B2H6 + 4H2O = 2H3BO4 + 4H2

this is now balanced

now... get moles of B2H6

MW of B2H6 = 27.66

mol = mass/MW = 76.99/27.66 = 2.783 mol of B2H6

raitos:

B2H6 + 4H2O = 2H3BO4 + 4H2

B2H6 --> 2 mol of H3BO4

2.783 mol --> 2*2.783 = 5.566 mol

mass = mol*MW = 5.566*77.8324 = 433.2 g of H3BO3

for H2:;

1 mol ofB2H6 = 4 mol of H2

2.783 mol --> 4*2.783 = 11.132 mol of H2

mass = mol*MW = 11.132*2 = 22.264 g