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Chapter 3: Problem 62

Egyptian Cosmetics \(\mathrm{Pb}\) (OH) Cl, one of the lead compounds used in ancient Egyptian cosmetics, was prepared from PbO according to the following recipe:$$\mathrm{PbO}(s)+\mathrm{NaCl}(a q)+\mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow \mathrm{Pb}(\mathrm{OH}) \mathrm{Cl}(s)+\mathrm{NaOH}(a q)$$.How many grams of \(\mathrm{PbO}\) and how many grams of \(\mathrm{NaCl}\) would be required to produce \(10.0 \mathrm{g}\) of \(\mathrm{Pb}(\mathrm{OH}) \mathrm{Cl} ?\)

Short Answer

Expert verified
Answer: To produce 10.0g of Pb(OH)Cl, we would need 8.59g of PbO and 2.25g of NaCl.

Step by step solution

01

Calculate the moles of Pb(OH)Cl

First, we need to find the molar mass of \(\mathrm{Pb(OH)Cl}\). Molar mass of \(\mathrm{Pb(OH)Cl} = \mathrm{Pb} + \mathrm{O} + \mathrm{H} + \mathrm{Cl} = 207.2\,\text{g/mol}+16.0\,\text{g/mol}+1.0\,\text{g/mol}+35.5\,\text{g/mol}=259.7\,\text{g/mol}\) Now, we can calculate the number of moles of \(\mathrm{Pb(OH)Cl}\) in \(10.0\,\mathrm{g}\): Moles of \(\mathrm{Pb(OH)Cl} = \dfrac{\text{mass}}{\text{molar mass}} = \dfrac{10.0\,\mathrm{g}}{259.7\,\mathrm{g/mol}} = 0.0385\,\mathrm{mol}\).
02

Determine moles of PbO and NaCl using stoichiometry

According to the balanced chemical equation, we have: 1 mole of \(\mathrm{PbO}\) reacts with 1 mole of \(\mathrm{NaCl}\) to produce 1 mole of \(\mathrm{Pb(OH)Cl}\). Thus, the number of moles of \(\mathrm{PbO}\) and \(\mathrm{NaCl}\) required will be equal to the number of moles of \(\mathrm{Pb(OH)Cl}\). Moles of PbO = Moles of NaCl = Moles of Pb(OH)Cl = 0.0385 mol
03

Calculate the mass of PbO and NaCl

Now, we can find the mass of \(\mathrm{PbO}\) and \(\mathrm{NaCl}\) required to produce \(10.0 \mathrm{g}\) of \(\mathrm{Pb(OH)Cl}\). Mass of \(\mathrm{PbO} = \text{moles} \times \text{molar mass} = 0.0385\,\text{mol} \times 223.2\,\text{g/mol} = 8.59\,\text{g}\) Mass of \(\mathrm{NaCl} = \text{moles} \times \text{molar mass} = 0.0385\,\text{mol} \times 58.5\,\text{g/mol} = 2.25\,\text{g}\) To produce \(10.0 \mathrm{g}\) of \(\mathrm{Pb(OH)Cl}\), we would need \(8.59\,\mathrm{g}\) of \(\mathrm{PbO}\) and \(2.25\,\mathrm{g}\) of \(\mathrm{NaCl}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chemical Reactions
Chemical reactions involve the transformation of one or more substances into new products. In the given reaction, we start with lead(II) oxide (\(\mathrm{PbO}\)), sodium chloride (\(\mathrm{NaCl}\)), and water (\(\mathrm{H_2O}\)) as reactants. These reactants chemically interact to produce lead(II) hydroxide chloride (\(\mathrm{Pb(OH)Cl}\)) and sodium hydroxide (\(\mathrm{NaOH}\)). The equation provided for this reaction is: \[\mathrm{PbO}(s)+\mathrm{NaCl}(aq)+\mathrm{H}_{2}\mathrm{O}(\ell) \rightarrow \mathrm{Pb(OH)Cl}(s)+\mathrm{NaOH}(aq)\] This balanced equation is crucial because it tells us the proportion of reactants needed and the products formed. The coefficients indicate that one mole of Lead(II) oxide reacts with one mole of sodium chloride, indicating a one-to-one ratio, ensuring no excess reactants remain. Balanced equations are like recipes. They guide the amounts of each ingredient required for the desired product, ensuring the conservation of matter.
Molar Mass Calculations
Molar mass is a fundamental concept in stoichiometry. It represents the mass of one mole of a substance, measured in grams per mole (g/mol). To calculate stoichiometric quantities, knowing the molar mass of all reactants and products is essential. For \(\mathrm{Pb(OH)Cl}\), the molar mass is calculated as follows: - Lead (Pb): 207.2 g/mol - Oxygen (O): 16.0 g/mol - Hydrogen (H): 1.0 g/mol - Chlorine (Cl): 35.5 g/mol Adding them, we get a total molar mass of \(259.7\, \mathrm{g/mol}\). This value is crucial to convert between grams of \(\mathrm{Pb(OH)Cl}\) and moles, which is a necessary step to find the amount of \(\mathrm{PbO}\) and \(\mathrm{NaCl}\) needed. By dividing the given mass by the molar mass, we can determine that \(10.0\, \mathrm{g}\) of \(\mathrm{Pb(OH)Cl}\) corresponds to \(0.0385\, \mathrm{mol}\). Having accurate molar masses ensures precise calculations and is key to mastering stoichiometry.
Reactant Quantities
Determining the exact quantities of reactants required in a chemical reaction involves a process called stoichiometry. This method relies on the balanced chemical equation to relate the moles of reactants to the moles of products. In our specific reaction:
  • 1 mole \(\mathrm{PbO}\) produces 1 mole \(\mathrm{Pb(OH)Cl}\)
  • 1 mole \(\mathrm{NaCl}\) produces 1 mole \(\mathrm{Pb(OH)Cl}\)
This one-to-one stoichiometric ratio means that the moles of each reactant needed are equivalent to the moles of \(\mathrm{Pb(OH)Cl}\) produced. Since we calculated \(0.0385\, \mathrm{mol}\) of \(\mathrm{Pb(OH)Cl}\) is needed, the same amount of \(\mathrm{PbO}\) and \(\mathrm{NaCl}\) is required.Finally, to find the mass of each reactant, we multiply their number of moles by their respective molar masses:
  • \(\mathrm{PbO}: 0.0385\, \mathrm{mol} \times 223.2\, \mathrm{g/mol} = 8.59\, \mathrm{g}\)
  • \(\mathrm{NaCl}: 0.0385\, \mathrm{mol} \times 58.5\, \mathrm{g/mol} = 2.25\, \mathrm{g}\)
Quantifying reactants correctly is vital for the efficiency and success of chemical processes.

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Most popular questions from this chapter

Aluminum reacts with elemental oxygen at high temperatures to give pure aluminum oxide. What is the coefficient of \(\mathrm{O}_{2}(\mathrm{g})\) in the balanced chemical equation?

Gold can be extracted from the surrounding rock by using a solution of sodium cyanide. While effective for isolating gold, toxic cyanide finds its way into watersheds, causing environmental damage and harming human health.\(4 \mathrm{Au}(s)+8 \mathrm{NaCN}(a q)+\mathrm{O}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow\).\(4 \mathrm{NaAu}(\mathrm{CN})_{2}(a q)+4 \mathrm{NaOH}(a q)\) \(2 \mathrm{NaAu}(\mathrm{CN})_{2}(a q)+\mathrm{Zn}(s) \rightarrow 2 \mathrm{Au}(s)+\mathrm{Na}_{2}\left[\mathrm{Zn}(\mathrm{CN})_{4}\right](a q)\).a. If a sample of rock contains \(0.009 \%\) gold by mass, how much \(\mathrm{NaCN}\) is needed to extract the gold from 1 metric ton ( 1 metric ton \(=10^{3} \mathrm{kg}\) ) of rock as \(\mathrm{NaAu}(\mathrm{CN})_{2} ?\) b. How much zinc is needed to convert the \(\mathrm{NaAu}(\mathrm{CN})_{2}\) from part (a) to metallic gold? c. The gold recovered in part (b) is manufactured into a gold ingot in the shape of a cube. The density of gold is \(19.3 \mathrm{g} / \mathrm{cm}^{3} .\) How big is the cube of gold?

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