Question

4 g of hydrogen (${H_2}$ ), 64 g of sulphur (S) and 44.8 L of ${O_2}$ at STP reacts and forms ${H_2}S{O_4}$ . If 49 g of ${H_2}S{O_4}$ is formed, then $\%$ yield is?
A. 25 $\%$
B. 50 $\%$
C. 75 $\%$
D. 100 $\%$

Answer 1

In chemistry, compounds react to form new products but they do so in specific proportions. Different proportions of different reactions may lead to the formation of different products. Some laws govern the reaction and the proportions in which the compounds react and thus form a proportionate reaction.

Complete step by step answer:
S.T.P is the acronym for the standard temperature and pressure. It defines the state in which the temperature and pressure are defined and the volume taken by any gas is found to be 22.4 L.
1 mole of all the gases at S.T.P is found to acquire a volume of 22.4L.
The reaction which takes place in the above question will be
${H_2}\, + \,\,S\, + \,\,2{O_2}\,\, \to \,\,{H_2}S{O_4}$
The mass of the involved components are represented as

${H_2}$	$S$	${O_2}$	${H_2}S{O_4}$
2g	32g	x	98g

This reaction is the basic reaction going with minimum whole-number coefficients. The stoichiometric ratio in which the reaction takes place is depicted in the above reaction.
The molar ratio in which the reactants react with each other will be $H:S:O = 1:1:2$
So when the amount of oxygen is given to be 44.8 L we can deduce that 2 moles of oxygen is being reacted.
Since the question is based on the formation of 49 g of ${H_2}S{O_4}$ we can say that
1 mole ${H_2}S{O_4}$ is 98 g
0.5 mole ${H_2}S{O_4}$ is 49 g
So since only half mole of product is formed so according to the molar ratio the $\%$ yield will be calculated as
$\%$ yield = $\dfrac{{49}}{{98}} \times 100 = 50\%$

So, the correct answer is Option B.

Note: Charles law is a gas law which is derived after experimentation. It describes how gases tend to increase in volume when they are heated. The law states that, when the pressure of the gas is held constant, the kelvin temperature and the volume of the gas will be directly proportional. This law can be mathematically stated as
$\dfrac{{{V_1}}}{{{T_1}}} = \dfrac{{{V_2}}}{{{T_2}}}$