Calculate the mass (in grams) of hydrogen gas produced when 33.61 g of diborane (B2H) reacts with excess water.
B Holg) + 6H2O(1)→ 2 H3B03(s) + 6H2(g)
Only enter numerical value without units.

14.40g of H2

Explanation:

The balanced equation for the reaction is given below:

B2H6(g) + 6H2O(l) → 2H3BO3(s) + 6H2(g)

Next, we shall determine the mass of B2H6 that reacted and the mass of H2 produced from the balanced equation. This is illustrated below:

Molar mass of B2H6 = (2x11) + (6x1) = 28g/mol

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

Molar mass of H2 = 2x1 = 2g/mol

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

From the balanced equation above,

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

Finally, we shall determine the mass of H2 produced when 33.61g of B2H6 reacted with excess water. This can be obtained as follow:

From the balanced equation above,

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

Therefore, 33.61g of B2H6 will react to produce = (33.61 x 12)/28 = 14.40g of H2.

Therefore, 14.40g of H2 were produced from the reaction.

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