50 gm of CaCO3 is allowed to react with 68.6 gm of H3PO4 then select the correct option(s): 3CaCO3+2H3PO4⟶Ca3PO42+3H2O+3CO2

Book Online Demo

Check Your IQ

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

50 gm of ${CaCO}_{3}$ is allowed to react with 68.6 gm of $H_{3} {PO}_{4}$ then select the correct option(s): $3 {CaCO}_{3} + 2 H_{3} {PO}_{4} ⟶ {Ca}_{3} {({PO}_{4})}_{2} + 3 H_{2} O + 3 {CO}_{2}$

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

51.67 gm salt is formed

$n_{{CO}_{2}} = 0.5 moles$

Amount of unreacted reagent = 35.93 gm

0. 7 moles of ${CO}_{2}$ is evolved

answer is A, B, C.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

$3 {CaCO}_{3} + 2 H_{3} {PO}_{4} \to {Ca}_{3} {({PO}_{4})}_{2} + 3 H_{2} O + 3 {CO}_{2} \begin{array}{r} n = \frac{50}{100} \\ \underset{(LR)}{= 0.5} \end{array} \begin{array}{r} = \frac{68.6}{98} \\ 0.7 \end{array} - - - - \begin{array}{cc} [0.7 - \frac{0.5}{3} \times 2] \times 98 & \frac{0.5}{3} \times 310 \\ = 35.93 g & = 51.67 g \end{array} 0.5 mole$