Question

The volume of carbon dioxide gas evolved at S.T.P. by heating 7.3 gm of $Mg{(HC{O_3})_2}$ will be:
A.2240ml
B.1120ml
C.2340ml
D.2000ml

Answer 1

To solve such questions, we must know the balanced chemical equation of the reaction. We can then find out the amount of $Mg{(HC{O_3})_2}$ reacting in moles and this will also give us the amount of carbon dioxide formed. It will then be easy for us to calculate the volume of gas from the ideal gas equation at STP.

Formula used: $PV = nRT$ and
$moles = \dfrac{{given\,\,mass}}{{molar\,mass}}$

Complete step-by-step answer:
We can understand how the reaction happening such that magnesium bicarbonate on heating forms magnesium hydroxide and carbon dioxide. The balanced chemical equation for this reaction can be written as:
$Mg{(HC{O_3})_2}\mathop \to \limits^\Delta Mg{(OH)_2} \downarrow + 2C{O_2} \uparrow$
Let us now find the moles of $Mg{(HC{O_3})_2}$ taking part in the reaction. We have given mass of this compound as 7.3gms and we know the molar mass of this compound is 146.34gms
So, using the formula of moles and putting the values of given mass and molar mass in it, we get the moles of $Mg{(HC{O_3})_2}$ .

$$moles = \dfrac{{given\,\,mass}}{{molar\,mass}} \\\ moles = \dfrac{{7.3}}{{146.34}} \simeq 0.05moles \\\$$

Now, looking at the stoichiometry of the reactants and products in the chemical reaction, we see that 1 mole of $Mg{(HC{O_3})_2}$ produces 2 moles of carbon dioxide. Therefore, 0.05 moles of $Mg{(HC{O_3})_2}$ will produce 0.1 moles of carbon dioxide.
Let us now determine the volume of carbon dioxide present in the reaction using the ideal gas equation i.e. $PV = nRT$ where n is the number of moles of gas and R is the gas constant.
At STP, pressure is 1 atm and temperature is 273K. Putting these values in the ideal gas equation we get,

$$1atm \times V = 0.1moles \times 0.0821Latm/molK \times 273K \\\ \therefore V = 2.24litres\,\,or\,\,2240ml \\\$$

The volume of carbon dioxide gas evolved at S.T.P. by heating 7.3 gm of $Mg{(HC{O_3})_2}$ will be 2240ml.
Hence, the correct option is (A).

Note: There is another simple way of calculating the volume of carbon dioxide from the simplified gas equation at STP, which is $V = n \times 22.4$ . We know now that there are 0.1 moles of carbon dioxide present so n will be 0.1. substituting the value of n in $V = n \times 22.4$ , we get

$$V = 0.1 \times 22.4 \\\ \,V = 2.24L\,\,or\,\,2240ml \\\$$