How many moles of BCl3 are needed to produce 10.0 g of HCl(aq) in the following reaction? (HCl molar mass is 36.46 g/mol). BCl3(g) + 3 H2O(l) → 3 HCl(aq) + B(OH)3(aq)

The number of moles of BCl₃ needed is 0.0914 mole

From the question,

We are to determine the number of moles of BCl₃ needed to produce 10.0 g of HCl

From the balanced chemical equation,

BCl₃(g) + 3H₂O(l) → 3HCl(aq) + B(OH)₃(aq)

This means,

1 mole of BCl₃ reacts with 3 moles of H₂O to produce 3 moles of HCl and 1 mole of B(OH)₃

Now, we will determine the number of moles of HCl required to be produced

Mass of HCl = 10.0g

Using the formula

Molar mass of HCl = 36.46 g/mol

∴ Number of moles of HCl required =

Number of moles of HCl required = 0.27427 mole

Since,

1 mole of BCl₃ are needed to produce 3 moles of HCl

Then,

mole of BCl₃ would be needed to produce 0.27427 mole of HCl

∴ 0.0914 mole of BCl₃ would be needed to produce 10.0 g of HCl

Hence, the number of moles of BCl₃ needed is 0.0914 mole

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Moles of BCl₃ needed = 0.089 mol

Explanation:

Given data:

Moles of BCl₃ needed = ?

Mass of HCl produced = 10.0 g

Solution:

Chemical equation:

BCl₃ + 3H₂O  →     3HCl + B(OH)₃

Number of moles of HCl:

Number of moles = mass/molar mass

Number of moles = 10.0 g/ 36.46 g/mol

Number of moles = 0.27 mol

Now we will compare the moles of HCl with BCl₃.

               HCl             :           BCl₃

                 3               :             1

             0.27             :            1/3×0.27 = 0.089 mol