0.0075 mole of calcium trioxocarbonate (iv) is added to 0.015 mole of a solution of

Explanation

First, let's write down the balanced chemical equation for the reaction between calcium trioxocarbonate (IV) and hydrochloric acid (HCl):

\(CaCO_3 + 2HCl \rightarrow CaCl_2 + H_2O + CO_2\)

Now, we can determine the limiting reactant in this reaction. We have 0.0075 moles of calcium trioxocarbonate (IV) and 0.015 moles of HCl. According to the balanced equation, one mole of calcium trioxocarbonate (IV) reacts with two moles of HCl. So, we will divide the moles of each reactant by their respective stoichiometric coefficients:

\(\frac{0.0075 \, \text{moles} \, CaCO_3}{1} = 0.0075\)

\(\frac{0.015 \, \text{moles} \, HCl}{2} = 0.0075\)

Since both values are the same, the reaction goes to completion, and there is no limiting reactant.

To find the volume of the gas evolved at S.T.P (Standard Temperature and Pressure), we need to determine the moles of carbon dioxide (CO₂) produced. According to the balanced equation, one mole of calcium trioxocarbonate (IV) produces one mole of CO₂. Therefore, 0.0075 moles of calcium trioxocarbonate (IV) will produce 0.0075 moles of CO₂.

At S.T.P, one mole of any gas occupies a volume of 22.4 L (or 22400 cm³). To find the volume of 0.0075 moles of CO₂ gas evolved, we can use the following equation:

\(\text{Volume of gas} = \text{moles of gas} \times \text{molar volume at S.T.P}\)

\(\text{Volume of gas} = 0.0075 \, \text{moles} \times 22400 \, cm^3\, \text{per mole} = 168 \, cm^3\)

Thus, the volume of the gas evolved at S.T.P is168 cm³ (Option B).