Simple method to find limiting reactant when question is in percentage form

. \(20\text{ g}\) of \(40%\) pure \(\text{CaCO}_{3}\) if reacted with \(5\text{ g}\) of \(\text{HCl}\) to produce \(\text{CaCl}_{2}\), \(\text{H}_2\text{O}\) and \(\text{CO}_{2}\).a. Find which one is limiting reactant and why?b. Calculate mass of \(\text{CaCl}_{2}\) formed.c. How many number of water molecules are produced?d. Calculate the volume of \(\text{CO}_{2}\) produced at \(27^{\circ }\text{C}\) and \(0.5\text{ atm}\) pressure.Answer:(a) \(\text{HCl}\)(b) \(7.602\text{ g}\)(c) \(4.12 \times 10^{22}\text{ water molecules}\)(d) \(3.37\text{ litres}\)