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Chapter 4: Problem 96

Chalkboard chalk is made from calcium carbonate and calcium sulfate, with minor impurities such as \(\mathrm{SiO}_{2} .\) Only the \(\mathrm{CaCO}_{3}\) reacts with dilute \(\mathrm{HCl}(\mathrm{aq})\) What is the mass percent \(\mathrm{CaCO}_{3}\) in a piece of chalk if a 3.28 -g sample yields \(0.981 \mathrm{g} \mathrm{CO}_{2}(\mathrm{g}) ?\) $$\begin{aligned} \mathrm{CaCO}_{3}(\mathrm{s})+2 \mathrm{HCl}(\mathrm{aq}) & \longrightarrow \mathrm{CaCl}_{2}(\mathrm{aq}) +\mathrm{H}_{2} \mathrm{O}(\mathrm{l})+\mathrm{CO}_{2}(\mathrm{g}) \end{aligned}$$

Short Answer

Expert verified
The mass percent of \(\mathrm{CaCO}_{3}\) in the chalk sample is 68.0%.

Step by step solution

01

Calculate Molar Masses

The first step is to calculate the molar mass of \(\mathrm{CO}_{2}\) and \(\mathrm{CaCO}_{3}\). The molar mass of \(\mathrm{CO}_{2}\) is \(12.01 g/mol + 2(16.00 g/mol) = 44.01 g/mol\). The molar mass of \(\mathrm{CaCO}_{3}\) is \(40.08 g/mol + 12.01 g/mol + 3(16.00 g/mol) = 100.09 g/mol\).
02

Calculate Moles of Produced CO2

Next, calculate the number of moles of \(\mathrm{CO}_{2}\) produced from the reaction. This can be done by dividing the given mass of \(\mathrm{CO}_{2}\) by its molar mass. In this case, \(0.981 g ÷ 44.01 g/mol = 0.0223 mol\).
03

Calculate Mass of Reacted CaCO3

With the balanced chemical equation, we can conclude that since 1 mole of \(\mathrm{CaCO}_{3}\) results in 1 mole of \(\mathrm{CO}_{2}\), the moles of \(\mathrm{CaCO}_{3}\) reacted will be the same as the moles of \(\mathrm{CO}_{2}\) produced. Therefore, 0.0223 mol of \(\mathrm{CaCO}_{3}\) reacted. The mass can therefore be calculated as: \(0.0223 mol × 100.09 g/mol = 2.23 g\).
04

Calculate Mass Percent CaCO3

The mass percent of \(\mathrm{CaCO}_{3}\) in the chalk sample can be calculated using the formula: mass percent = mass of component in sample / total mass of sample × 100%. Therefore, the mass percent of \(\mathrm{CaCO}_{3}\) is \(2.23 g / 3.28 g × 100% = 68.0% .\)

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Most popular questions from this chapter

High-purity silicon is obtained using a three-step process. The first step involves heating solid silicon dioxide, \(\mathrm{SiO}_{2^{\prime}}\) with solid carbon to give solid silicon and carbon monoxide gas. In the second step, solid silicon is converted into liquid silicon tetrachloride, \(\mathrm{SiCl}_{4}\) by treating it with chlorine gas. In the last step, \(\mathrm{SiCl}_{4}\) is treated with hydrogen gas to give ultrapure solid silicon and hydrogen chloride gas. (a) Write balanced chemical equations for the steps involved in this three- step process. (b) Calculate the masses of carbon, chlorine, and hydrogen required per kilogram of silicon.

A reaction mixture contains \(1.0 \mathrm{mol} \mathrm{CaCN}_{2}\) (calcium cyanamide) and \(1.0 \mathrm{mol} \mathrm{H}_{2} \mathrm{O}\). The maximum number of moles of \(\mathrm{NH}_{3}\) produced is (a) \(3.0 ;\) (b) 2.0 (c) between 1.0 and 2.0; (d) less than 1.0. $$\mathrm{CaCN}_{2}(\mathrm{s})+3 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \longrightarrow \mathrm{CaCO}_{3}+2 \mathrm{NH}_{3}(\mathrm{g})$$

In the reaction shown, \(100.0 \mathrm{g} \mathrm{C}_{6} \mathrm{H}_{11} \mathrm{OH}\) yielded \(64.0 \mathrm{g}\) \(\mathrm{C}_{6} \mathrm{H}_{10} .\) (a) What is the theoretical yield of the reaction? (b) What is the percent yield? (c) What mass of \(\mathrm{C}_{6} \mathrm{H}_{11} \mathrm{OH}\) would produce \(100.0 \mathrm{g} \mathrm{C}_{6} \mathrm{H}_{10}\) if the percent yield is that determined in part (b)? $$\mathrm{C}_{6} \mathrm{H}_{11} \mathrm{OH} \longrightarrow \mathrm{C}_{6} \mathrm{H}_{10}+\mathrm{H}_{2} \mathrm{O}$$

Nitric acid, \(\mathrm{HNO}_{3}\), can be manufactured from ammonia, \(\mathrm{NH}_{3}\), by using the three reactions shown below. $$\begin{aligned} &\text { Step 1: 4 NH }_{3}(\mathrm{g})+5 \mathrm{O}_{2}(\mathrm{g}) \rightarrow 4 \mathrm{NO}(\mathrm{g})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})\\\ &\text { Step 2: } 2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{NO}_{2}(\mathrm{g})\\\ &\text { Step 3: } 3 \mathrm{NO}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightarrow 2 \mathrm{HNO}_{3}(\mathrm{aq})+\mathrm{NO}(\mathrm{g}) \end{aligned}$$ What is the maximum number of moles of \(\mathrm{HNO}_{3}\) that can be obtained from 4.00 moles of \(\mathrm{NH}_{3}\) ? (Assume that the NO produced in step 3 is not What is the maximum number of moles of \(\mathrm{HNO}_{3}\) that can be obtained from 4.00 moles of \(\mathrm{NH}_{3}\) ? (Assume that the NO produced in step 3 is not recycled back into step 2.) (a) 1.33 mol; (b) 2.00 mol; (c) 2.67 mol; (d) 4.00 mol; (e) 6.00 mol.

What is the molarity of \(\mathrm{NaCl}(\mathrm{aq})\) if a solution has 1.52 ppm Na? Assume that NaCl is the only source of Na and that the solution density is \(1.00 \mathrm{g} / \mathrm{mL}\) (The unit \(p p m\) is parts per million; here it can be taken to mean g Na per million grams of solution.)
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