What mass of CaCO3 is required to react completely with 25 ml of 0.75 M HCl according to the reaction CaCO3(s)+2HCl(aq) arrow CaCl2(aq)+CO2(g)+H2O(l)

Question

What mass of CaCO3 is required to react completely with 25 ml of 0.75 M HCl according to the reaction CaCO3(s)+2HCl(aq) arrow CaCl2(aq)+CO2(g)+H2O(l) (A) 1 g (B) 0.3 g (C) 0.8 g (D) 0.93 g

Tardigrade Admin · Accepted Answer

CaCO 3( s )+2 HCl ( aq ) arrow CaCl 2( aq )+ CO 2( g )+ H 2 O ( l ) Moles present in 25 ml of 0.75 M HCl =(0.75/ l 000) × 25 =0.01875 moles Since 2 moles react completely with CaCO 3=100 g 0.01875 moles react completely with CaCO 3=(100/2) × 0.01875 =0.9375 g

Q. What mass of CaCO3​ is required to react completely with 25 ml of 0.75 M HCl according to the reaction CaCO3​(s)+2HCl(aq)→CaCl2​(aq)+CO2​(g)+H2​O(l)

1 g

0.3 g

0.8 g

0.93 g

CaCO3​(s)+2HCl(aq)→CaCl2​(aq)+CO2​(g)+H2​O(l) Moles present in 25ml of 0.75MHCl=l0000.75​×25 =0.01875 moles Since 2 moles react completely with CaCO3​=100g 0.01875 moles react completely with CaCO3​=2100​×0.01875 =0.9375g

Q. What mass of $C a C O_{3}$ is required to react completely with 25 ml of 0.75 M HCl according to the reaction $C a C O_{3} (s) + 2 H Cl (a q) \to C a C l_{2} (a q) + C O_{2} (g) + H_{2} O (l)$