Question

toilet bowl cleaners often contain hydrochloric acid to dissolve the calcium carbonate deposits that accumulate within a toilet bowl.
part a
how much calcium carbonate in grams can be dissolved by 5.4 g of hcl? (hint: begin by writing a balanced equation for the reaction between hydrochloric acid and calcium carbonate)
express your answer in grams to two significant figures.
mass =

Explanation:

Step1: Write the balanced chemical equation

The reaction between hydrochloric acid ($HCl$) and calcium carbonate ($CaCO_3$) is:
$$CaCO_{3}+2HCl = CaCl_{2}+H_{2}O + CO_{2}\uparrow$$

Step2: Calculate the molar mass

The molar mass of $HCl$ is $M_{HCl}=1 + 35.5=36.5\space g/mol$.
The molar mass of $CaCO_{3}$ is $M_{CaCO_{3}}=40 + 12+3\times16 = 100\space g/mol$.

Step3: Calculate the number of moles of $HCl$

Given mass of $HCl$, $m_{HCl}=5.4\space g$.
Number of moles of $HCl$, $n_{HCl}=\frac{m_{HCl}}{M_{HCl}}=\frac{5.4}{36.5}\space mol$.

Step4: Use stoichiometry to find moles of $CaCO_{3}$

From the balanced equation, the mole ratio of $CaCO_{3}$ to $HCl$ is $1:2$.
So, number of moles of $CaCO_{3}$, $n_{CaCO_{3}}=\frac{1}{2}n_{HCl}=\frac{1}{2}\times\frac{5.4}{36.5}\space mol$.

Step5: Calculate the mass of $CaCO_{3}$

Mass of $CaCO_{3}$, $m_{CaCO_{3}}=n_{CaCO_{3}}\times M_{CaCO_{3}}$
$m_{CaCO_{3}}=\frac{1}{2}\times\frac{5.4}{36.5}\times100\space g$
$m_{CaCO_{3}}=\frac{540}{73}\space g\approx7.4\space g$

Answer:

$7.4\space g$