Question

How much \(POC{l_3}\) can form from \(25.0\;g\)of \(PC{l_5}\) and the appropriate amount of \({H_2}O\) ?

Short answer

The Mass of \(POC{l_3}\) created is \(18.4\;g\).

Step-by-step solution

Definition of POCl3

• Phosphoryl chloride has the formula POCl₃ and is a colourless liquid. In damp air, it hydrolyzes, generating phosphoric acid and hydrogen chloride fumes.
• It is made on a big scale in an industrial setting from phosphorus trichloride and oxygen or phosphorus pentoxide.

Determine the \(POC{l_3}\) can form from 25.0g of \(PC{l_5}\) and the appropriate amount of \({H_2}O\)

Given:

Mass of \(PC{l_5} = 25.0\;g\)

The reaction of \(PC{l_5}\) with \({H_2}O\) can be expressed as follows:

\(PC{l_5} + {H_2}O \to POC{l_3} + 2HCl\)

Based on the stoichiometry of the reaction:

1 mole of \(PC{l_5}\) 1 mole of a substance is combined with 1 mole of a substance \({H_2}O\) to make one mole of \(POC{l_3}\)

Moles of \(PC{l_5}\)reacted \( = \frac{{25.0\;g}}{{208.24\;g - mo{l^1}}} = 0.1201\) moles

Based on the stoichiometry of the reaction:

Moles of \(POC{l_3}\)formed =0.1201 moles

Mass of \(POC{l_3}\) formed =0.1201 moles \( \times 153.33\;g \times mo{l^{ - 1}} = 18.415\;g\)

Around a pound of ammonia is created \(18.4\;g\).