Question

In the reaction:
$N{{a}_{2}}C{{O}_{3}}+2HCl\xrightarrow{{}}2NaCl+C{{O}_{2}}+{{H}_{2}}O$
If 106.0g of $N{{a}_{2}}C{{O}_{3}}$ reacts with 109.5g of $HCl$. Which of the following are correct?
A. $HCl$ is in excess
B. 117.0g of $NaCl$ is formed
C. The volume of $C{{O}_{2}}$ produced at 1 bar and 273K is 22.4L
D. The volume of $C{{O}_{2}}$ produced at 1 bar and 298K is 22.4L

Answer 1

Calculate the limiting reagent by the molecular masses and stoichiometric coefficients of the reaction and then find amount and volume of product using mole concept.

Complete answer:
To find the limiting reagent in this reaction, we will need to divide the given number of moles of the reactant by their stoichiometric coefficients given in the rection. The reactant that has the lesser value among these will be the limiting reagent.
Now, to calculate the number of moles of reactants given we need to find their molecular masses.
$\text{Molecular mass of }N{{a}_{2}}C{{O}_{3}}=(2\times 23)+(1\times 12)+(3\times 16)$
$\text{Molecular mass of }N{{a}_{2}}C{{O}_{3}}=106g/mol$
$\text{Molecular mass of }HCl=(1\times 1)+(1\times 35.5)$
$\text{Molecular mass of }HCl=36.5g/mol$
The number of moles of reactants present can be calculated by dividing the given mass by the molecular mass.
$\text{Number of moles of }N{{a}_{2}}C{{O}_{3}}=\dfrac{106g}{106g/mol}$
$\text{Number of moles of }N{{a}_{2}}C{{O}_{3}}=1mol$
$\text{Number of moles of }HCl=\dfrac{109.5g}{36.5g/mol}$
$\text{Number of moles of }HCl=3mol$
Now, to find the limiting reagent:
For $HCl$= 3/2
For $HCl$= 1.5
For $N{{a}_{2}}C{{O}_{3}}$= 1/1
For $N{{a}_{2}}C{{O}_{3}}$= 1
Thus $N{{a}_{2}}C{{O}_{3}}$ has the lesser ratio and it is the limiting reagent. This implies that $HCl$ will be present in excess.
Hence, ‘A. $HCl$ is in excess’ is true.
Now, according to the reaction, 1mol of $N{{a}_{2}}C{{O}_{3}}$ produces 2mol of $NaCl$. The weight of 2mol of $NaCl$ can be calculated using the molecular weight.
$\text{Molecular weight of }NaCl=(1\times 23)+(1\times 35.5)$
$\text{Molecular weight of }NaCl=58.5g/mol$
Thus, the weight of 2mol of $NaCl$is:
$\text{Weight of }NaCl=(58.5g/mol)\times (2mol)$
$\text{Weight of }NaCl=117g$
Hence, ‘B. 117.0g of $NaCl$is formed’ is true.
The amount of $C{{O}_{2}}$ produced by 1mol of $N{{a}_{2}}C{{O}_{3}}$ is 1 mol.
The conditions given in ‘C. The volume of $C{{O}_{2}}$ produced at 1 bar and 273K is 22.4L’ are the conditions present a STP (Standard Temperature Pressure) i.e. temperature is 273K and pressure is 1bar.
It is postulated that the volume of one mole of any gas at STP is always 22.4L.
Hence, ‘C. The volume of $C{{O}_{2}}$ produced at 1 bar and 273K is 22.4L’ is true.
Now considering the last option ‘D. The volume of $C{{O}_{2}}$ produced at 1 bar and 298K is 22.4L’ the pressure is the same as the previous option, but the temperature is increased. We know that when the temperature increases, the volume of the gas will also increase. Thus, at a higher temperature the volume of the gas will be greater than it was at STP i.e. it will be greater than 22.4L.

Hence, ‘D. The volume of $C{{O}_{2}}$ produced at 1 bar and 298K is 22.4L’ is false.

Note: Please note that while calculating the weight and the number of moles of the product, always consider the number of moles of the limiting reagent that is required (Here, 1mol $N{{a}_{2}}C{{O}_{3}}$). Considering the number of moles of the reagent in excess (Here, 3mol $HCl$) will yield wrong results since the limiting reagent will be exhausted and the reaction will not occur.