Be sure to answer all parts. Calculate the mass of each product formed when 70.76 g of diborane (B2H6) reacts with excess water: B2H6(g) + H2O(l) → H3BO3(s) + H2(g) [unbalanced] Mass of H3BO3: g Mass of H2: g

A. 313.37 g of H3BO3.

B. 30.33 g of H2.

Explanation:

We'll begin by writing a balanced equation for the reaction. This is given below:

B2H6 + 6H2O → 2H3BO3 + 6H2

Next, we shall determine the mass of B2H6 that reacted and the masses of H3BO3 and H2 produced from the balanced equation equation. This can be obtained as follow:

Molar mass of B2H6 = (2×11) + (1×6) = 22 + 6 = 28 g/mol

Mass of B2H6 from the balanced equation = 1 × 28 = 28 g

Molar mass of H3BO3 = (1×3) + 11 + (3×16) = 3 + 11 + 48 = 62 g/mol

Mass of H3BO3 from the balanced equation = 2 × 62 = 124 g

Molar mass of H2 = 2×1 = 2 g/mol

Mass of H2 from the balanced equation = 6 × 2 = 12 g

From the balanced equation above:

28 g of B2H6 reacted to produce 124 g of H3BO3 and 12 g of H2.

A. Determination of the mass of H3BO3 produced from the reaction.

This can be obtained as follow:

From the balanced equation above,

28 g of B2H6 reacted to produce 124 g of H3BO3.

Therefore, 70.76 g of B2H6 will react to produce = (70.76 × 124)/28 = 313.37 g of H3BO3.

Therefore, 313.37 g of H3BO3 were obtained from the reaction.

B. Determination of the mass of H2 produced from the reaction.

This can be obtained as follow:

From the balanced equation above,

28 g of B2H6 reacted to produce 12 g of H2.

Therefore, 70.76 g of B2H6 will react to produce = ((70.76 × 12)/28 = 30.33 g of H2.

Therefore, 30.33 g of H2 were obtained from the reaction.