Question

A 1.248 g sample of limestone rock is pulverized and then treated with 30.00 ml of 1.035 m hcl solution. the excess acid then requires 11.56 ml of 1.010 m naoh for neutralization. part a calculate the percent by mass of calcium carbonate in the rock, assuming that it is the only substance reacting with the hcl solution.

Answer 1

First, we will find the moles of HCl entered. This will require the use of the formula:

Moles = molarity * volume (in L)
Moles = 1.035 * 0.03
Moles of HCl = 0.0315

Next, we find the amount reacted with NaOH. Note that each mole of NaOH reacts with one mole of HCl (determined from reaction equation)

Moles of NaOH = 1.01 * 0.01156
Moles of NaOH = 0.01168

Moles of HCl reacted with calcium carbonate = 0.0315 - 0.01168
= 0.01982 mole

The reaction equation is:

CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O

So the moles of CaCO₃ were half that of HCl, which means:
0.01982 / 2 = 0.00991

The mass of calcium carbonate is:

Mass = moles * molecular weight
Mass = 0.00991 * 100
Mass = 0.991 grams

Percentage mass = (0.991 / 1.248) * 100

The ore has 79.41% calcium carbonate by mass.

Answer 2

The percentage by mass of calcium carbonate present in rock is $\boxed{{\mathbf{77}}{\mathbf{.62 \% }}}$.

Further Explanation:

First, we have to find the excess number of moles of HCl acid that are neutralized by NaOH.

The number of moles of NaOH in 11.56 ml of 1.010 M NaOH solution is calculated as follows:

$\begin{aligned}{\text{Number of moles of NaOH}}\left({{\text{mol}}}\right)&={\text{Concentration }}\left( {{\text{mol/L}}}\right) \times{\text{Volume }}\left({\text{L}} \right)\\&= 1.010{\text{ mol/L}}\left({11.56{\text{ ml}}\times \frac{{1{\text{ L}}}}{{1000{\text{ml}}}}}\right)\\&=0.011676{\text{ mol}}\\\end{aligned}$

The balanced chemical reaction of NaOH and HCl is as follows:

${\text{NaOH}}\left({aq}\right)+{\text{HCl}}\left({aq}\right)\to{\text{NaCl}}\left({aq} \right) +{{\text{H}}_{\text{2}}}{\text{O}}\left( l \right)$

Since NaOH and HCl are reacted in 1:1 ratio, therefore, the excess number of moles of HCl are equal to number of moles of NaOH that is 0.011676 mol.

Now, we have to find how many moles of HCl initially reacted with limestone.

The initial number of moles of HCl in 30.00 ml of 1.035 M HCl solution is calculated as follows:

$\begin{aligned}{\text{Number of moles of HCl}}\left( {{\text{mol}}}\right)&={\text{Concentration }}\left( {{\text{mol/L}}}\right) \times {\text{Volume }}\left({\text{L}} \right)\\&= 1.035{\text{ mol/L}} \times \left( {30.00{\text{ ml}}\times \frac{{1{\text{ L}}}}{{1000\,{\text{ml}}}}}\right)\\&=0.03105{\text{ mol}}\\\end{aligned}$

Therefore, the number of moles of HCl initially reacted with limestone is calculated as follows:

$\begin{aligned}{\text{Number of moles of HCl reacted with CaC}}{{\text{O}}_3}&=\left({0.03105{\text{ mol}} - 0.011676{\text{ mol}}}\right)\\&={\text{0}}{\text{.019374 mol of HCl}}\\\end{aligned}$

Therefore, the number of moles of HCl initially reacted with limestone is 0.019374 mol.

The balanced chemical equation for the reaction of limestone $\left( {{\text{CaC}}{{\text{O}}_{\text{3}}}}\right)$ with HCl is as follows:

${\text{CaC}}{{\text{O}}_3}\left( s \right) + 2{\text{HCl}}\left( {aq} \right)\to{\text{C}}{{\text{O}}_{\text{2}}}\left( g \right) + {{\text{H}}_{\text{2}}}{\text{O}}\left( l \right) + {\text{CaC}}{{\text{l}}_2}\left( {aq} \right)$

The balanced chemical equation shows that 1 mole of ${\text{CaC}}{{\text{O}}_3}$ reacted with 2 moles of HCl to neutralize the reaction completely, therefore, the number of moles of  ${\text{CaC}}{{\text{O}}_3}$ neutralized by 0.019374 mol of HCl are calculated as follows:

$\begin{aligned}{\text{Amount of CaC}}{{\text{O}}_3}\left( {{\text{mol}}}\right)&= \left( {{\text{0}}{\text{.019374 mol of HCl}}}\right)\left({\frac{{1{\text{ mol CaC}}{{\text{O}}_3}}}{{{\text{2 mol HCl}}}}}\right)\\&=0.009687{\text{ mol CaC}}{{\text{O}}_3}\\\end{aligned}$

The molar mass of ${\text{CaC}}{{\text{O}}_3}$  is 100.0 g/mol.

Mass of 0.009687 mol of ${\text{CaC}}{{\text{O}}_3}$ is calculated as follows:

$\begin{aligned}{\text{Mass}}\left( {\text{g}}\right)&={\text{Number of moles}} \times{\text{Molarmass}}\left({{\text{g/mol}}}\right)\\&=0.009687{\text{mol}}\times{\text{100}}{\text{.0 g/mol}}\\&=0.9687{\text{g}}\\\end{aligned}$

The percentage by mass can be calculated as follows:

$\begin{aligned}{\text{Percent by mass}}\left( \%\right)&=\frac{{{\text{Mass of CaC}}{{\text{O}}_3}}}{{{\text{Mass of lime stone}}}}\times 100\\&=\frac{{0.9687{\text{ g}}}}{{1.248{\text{ g}}}}\times 100\\&=77.62{\text{ }}\%\\\end{aligned}$

Learn more:

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2. Calculation of moles of NaOH: https://brainly.com/question/4283309

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Mole concept

Keywords: Percentage by mass, calcium carbonate in rock, number of moles of HCl, excess number of moles, CaCO3, balance chemical equation, limestone.