Mole ConceptHard

Question

Diborane tetrachloride was treated with NaOH and the following reaction occurred.

B2Cl4 + NaOH → NaBO2 + H2O + H2 + NaCl

If 1362 ml of hydrogen gas is formed at STP, then how much B2Cl4 was consumed? (B = 11)

Options

A.9.97 g
B.9.84 g
C.0.0968 g
D.23.57 g

Solution

$B_{2}Cl_{2} + 6NaOH \rightarrow 2NaBO_{2} + 2H_{2}O + H_{2} + 4NaCl$

164 g 22700 mL at STP

∴ Mass of B2Cl4 needed =$\frac{164}{22700} \times 1362 = 9.84\text{ g}$

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