Consider the above reaction, what mass of will be formed if 250 mL of 0.76 M HCl reacts with 1000 g of ? (Given: Molar mass of Ca, C, O, H and Cl are 40, 12, 16, 1 and 35.5 g , respectively) [2025]

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
${CaCO}_{3} (s) + 2 HCl (aq) \to {CaCl}_{2} (aq) + {CO}_{2} (g) + H_{2} O (l)$

Consider the above reaction, what mass of ${CaCl}_{2}$ will be formed if 250 mL of 0.76 M HCl reacts with 1000 g of ${CaCO}_{3}$

(Given: Molar mass of Ca, C, O, H and Cl are 40, 12, 16, 1 and 35.5 g ${mol}^{- 1}$ , respectively) [2025]

1 3.908 g

2 2.636 g

3 10.545 g

4 5.272 g

Ans.
(3)

Moles of ${CaCO}_{3}$ $= \frac{{Mass of CaCO}_{3}}{{Molar mass of CaCO}_{3}} = \frac{1000}{100} mol = 10 mol$

$Moles of HCl = Molarity \times Volume in L = 0.76 M×0.25 L = 0.19 mol$

${CaCO}_{3} + 2 HCl \to {CaCl}_{2} + {CO}_{2} + H_{2} O$

As per reaction stoichiometry, 20 mol of HCl are needed to react with 10 mol ${CaCO}_{3}$ , but only 0.19 mol of HCl is available. Thus, HCl is the limiting reagent. Moles of ${CaCl}_{2}$ formed depends upon moles of limiting reagent i.e. HCl.

As per reaction stoichiometry, ${Moles of CaCl}_{2} (n_{{CaCl}_{2}}) = \frac{n_{HCl}}{2} = \frac{0.19}{2} mol$

${Mass of CaCl}_{2} = {Moles of CaCl}_{2} \times {Molar mass of CaCl}_{2} = \frac{0.19}{2} \times 111 g = 10.545 g$

Want Us to Email you About Special Offers & Updates?