Question

Chlorine gas is prepared by reaction of ${H_2}S{O_4}$ with $Mn{O_4}$ and $NaCl$. What volume of $C{l_2}$ will be produced at STP if $50g$ of $NaCl$ is taken in the reaction?A.$1.915L$B.$22.4L$C.$11.2L$D.$9.57L$

Hint: At first think about the reaction between ${H_2}S{O_4},Mn{O_2}$ and sodium chloride and balance the equation. From the balanced chemical reaction you can get some idea about the amount of chlorine gas produced from the reaction.

The overall chemical reaction of ${H_2}S{O_4},Mn{O_2},NaCl$ is,
$2NaCl + 3{H_2}S{O_4} + Mn{O_2} \to 2NaHS{O_4} + MnS{O_4} + C{l_2} + 2{H_2}O$
Molar mass of sodium chloride is $58.5g$
From the above balanced chemical reaction two moles of sodium chloride produces one mole of chlorine gas. So, the mass of two moles of sodium chloride is $117g$.
Standard pressure and temperature is the full form of STP. STP is defined as a temperature of $273.15K$ and an absolute pressure of exactly ${10^5}Pa$. The molar volume of gases around STP and at atmospheric pressure can be calculated by using the ideal gas equation. From the ideal gas law we get molar volume of a gas is $22.4$ litres at STP.
At STP one mole of a gas contains $22.4L$ of volume. So $117g$ of sodium chlorine produces $22.4L$ of chlorine gas. Then,
$50g$ of sodium chloride produces,
$= \dfrac{{22.4}}{{117}} \times 50 = 9.57L$ chlorine gas at STP.
The required volume of chlorine gas is $9.57L$.
Additional Information:-STP should not be confused with the standard state commonly used in thermodynamic evaluations of Gibbs energy of a reaction. At NTP(normal temperature and pressure) the temperature is $293.15K$ and the absolute pressure is $1atm$.