Chapter 5, Problem 5.55P

Interpretation Introduction

Interpretation:

Grams of phosphine $\left({\text{PH}}_{\text{3}}\right)$ formed when $37.5\text{g}$ of phosphorous and $83.0\text{L}$ of hydrogen gas reacts at STP is to be calculated.

Concept introduction:

Ideal gas law equation is as follows:

$PV=nRT$        (1)

Here,

$P$ is the pressure of the gas.

$V$ is the volume occupied by gas.

$n$ is the number of moles of gas.

$T$ is the temperature in Kelvin.

$R$ is the universal gas constant.

Rearrange equation (1) for $n$ as follows:

$n=\frac{PV}{RT}$        (2)

The formula to calculate no. of moles is as follows:

$n=\frac{\text{weight}\left(m\right)}{\text{molecular}\text{weight}\left(M\right)}$        (3)

The temperature and pressure at STP is $273\text{K}$ and $1\text{atm}$ respectively.

The reactant present in the lower quantity of moles is the limiting reagent.

When the number of atoms of each element in the reactant side is equal to number of atoms in the product side then the reaction is known as a balanced chemical reaction.

The reagent that is completely used or consumed in a chemical reaction is called a limiting reagent. The amount of product formed in any chemical reaction is in accordance with the amount of the limiting reagent of the reaction.