Chapter 7: Q11P (page 158)

Limestone consists mainly of the mineral calcite, ${CaCO}_{3}$ .The carbonate content of 0.5413 g
powdered limestone was measured by suspending the powder in water, adding 10.00 mL of $1.396 MHCl$ and heating to dissolve the solid and expel role="math" localid="1654937947195" ${CO}_{2} {CaCO}_{3} (s) + 2 H^{+} \to {Ca}^{2 +} + {CO}_{2} ↑ + H_{2} O$ Calcium carbonate FM 100.087
The excess acid required $39.96 mL$ of $0.1004 MNaOH$ for complete titration to a phenolphthalein end point. Find the weight percent of calcite in the limestone.

Short Answer

Expert verified

The powdered limestone contains $92.3 %$ calcite.

Step by step solution

Define the formula to find the weight percentage.

$The weight percent % w = \frac{mass of}{mass of solution} - \times 100$

Step 2:Calculate the number of moles of that reacted with CaCO3

$Given information mass of powdered limestone = 0.5413 g volume of 1.396 MJHCI = 10.00 mL = 0.01 L volume of 0.1004 MNaOHadded to neutralize the solution = 39.96 mL = 0.0396 L FM of {CaCO}_{3} = 100.087 g / mol Now calculate the mole of HCL . mole of reacted with {CaCO}_{3} = original mole of HCI - mole of excess HCI = (0.01 L (1.396 \frac{molHCI}{L}) - (0.0396 L) (0.1004 \frac{molHCI}{L}) = 9.98416 \times 10^{- 3} molHCI$

Calculate the mass of CaCO3.

Using mole concept

$mass of {CaCO}_{3} = 9.98416 \times 10^{- 3} molHCI (\frac{1 {molCaCO}_{3}}{2 mollICI}) (\frac{100.087 {gCaCO}_{3}}{1 molCaCO}) = 0.4996 {gCaCO}_{3} The weight percent of calcite in limestone . % w / w = \frac{mass of}{mass of solute} - \times 100 = \frac{0.4996 {gCaCO}_{3}}{0.5413 g {CaCO}_{3}} \times 100 = 92.3 % Thus, the powdered limestone contains 92.3 % calcite .$