Question

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

Answer 1

Answer: 937.5

Given

${\mathrm{CaCO}}_3(\mathrm{s})+2\mathrm{HCl}(\mathrm{aq})\to {\mathrm{CaCl}}_2(\mathrm{aq})+{\mathrm{CO}}_2(\mathrm{g})+{\mathrm{H}}_2\mathrm{O}(\mathrm{l})$

Moles present in 25 ml of 0.75 M HCl $=\frac{0.75}{1000}\times 25$

$=0.01875$ moles

Since 2 moles react completely with $\text{1}\text{mole}{\text{CaCO}}_{\text{3}}\text{=}\text{100}\text{g}$

0.01875 moles react completely with ${\text{CaCO}}_{\text{3}}=\frac{100}{2}\times 0.01875$

$0.9375\text{g}=937.5\text{mg}$