Question

What mass of $CaC{O}_3$ is required to react completely with 25 ml of 0.75 M HCl according to the reaction $CaC{O}_3\left(s\right)+2HCl\left(aq\right)\to CaC{l}_2\left(aq\right)+C{O}_2\left(g\right)+H_{2}O\left(l\right)$

• A. 1 g
• B. 0.3 g
• C. 0.8 g
• D. 0.93 g

Answer 1

Answer: (D)

$CaC{O}_3(s)+2HCl(aq)\to CaC{l}_2(aq)+C{O}_2(g)+H_{2}O(l)$

Moles present in $25ml$ of $0.75MHCl=\frac{0.75}{l000}\times 25$

$=0.01875$ moles

Since 2 moles react completely with $CaC{O}_3=100g$

0.01875 moles react completely with $CaC{O}_3=\frac{100}{2}\times 0.01875$

$=0.9375g$