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Given that, $1.24{\text{g P}}$ is present in $2.2{\text{g}}$ of:
A. ${P_2}{S_4}$
B. ${P_2}{S_2}$
C. ${P_4}{S_3}$
D. $P{S_2}$

Answer
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Hint: To answer this question, you must recall the basic laws of stoichiometry. We need to find the mass of phosphorus in each compound and then calculate its ratio to the molar mass of the compound. The compound which gives the ratio equal to $1.24/2.2$ will give us the answer.

Complete step by step solution:
We know that the mass of phosphorus is 31g and that of Sulphur is 32g.
Calculating the amount of phosphorous in each of the compound in the given options:
Considering option A, we have ${P_2}{S_4}$.
The molar mass of this compound is given as$ = 2 \times 31 + 4 \times 32 = 190$ and the mass of phosphorus in the compound $ = 2 \times 31 = 62$
Ratio is given by $\dfrac{{62}}{{190}} \ne \dfrac{{1.24}}{{2.2}}$.
Thus, this is incorrect.
Considering option B, we have ${P_2}{S_2}$.
The molar mass of this compound is given as$ = 2 \times 31 + 2 \times 32 = 126$ and the mass of phosphorus in the compound $ = 2 \times 31 = 62$
Ratio is given by $\dfrac{{62}}{{126}} \ne \dfrac{{1.24}}{{2.2}}$.
Thus, this is incorrect.
Considering option C, we have ${P_4}{S_3}$.
The molar mass of this compound is given as $ = 4 \times 31 + 3 \times 32 = 220{\text{g}}$ and the mass of phosphorus in the compound$ = 4 \times 31 = 124$ .Thus, we can say that 220 grams of ${P_4}{S_3}$ contains 124 grams of
Ratio is given by $\dfrac{{124}}{{220}} = \dfrac{{1.24}}{{2.2}}$.
Hence, $1.24{\text{g P}}$ is present in $2.2{\text{g}}$ of ${P_4}{S_3}$.
Option C is the correct answer.
Considering option D, we have $P{S_2}$.
The molar mass of this compound is given as$ = 31 + 2 \times 32 = 95$ and the mass of phosphorus in the compound$ = 31$
Ratio is given by $\dfrac{{31}}{{95}} \ne \dfrac{{1.24}}{{2.2}}$
Thus, this is incorrect.

Hence, the correct answer is C.

Note: Stoichiometry is based upon the very basic laws of chemistry that help to understand it better, namely, the law of conservation of mass, the law of definite proportions (the law of constant composition), the law of reciprocal proportions and the law of multiple proportions .
In general, different chemicals combine in definite ratios in chemical reactions. Since matter can neither be created nor destroyed, thus the amount of each element must be the same throughout the entire reaction.