Question

56 gm of CaO has been mixed with 63 gm of HNO₃, the amount of Ca(NO₃)₂ formed is

(a) 4 gm

(b) 3.28 gm

(c) 164 gm

(d) 82 gm

Answer 1

Correct option is (d) 82 gm

CaO + 2HNO₃ → Ca(NO₃)₂ + H₂O

Moles of CaO = \(\frac{56}{56}\) = 1; Moles of HNO₃ = \(\frac{63}{63}\) = 1; HNO₃ is limiting reagent.

Hence, moles of Ca(NO₃)₂ = \(\frac 12\); Wt. of Ca(NO₃)₂ = \(\frac 12\) × 164 = 82 g