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Chapter 4: Problem 83

How ma475.15 grams \(\mathrm{HCl}\)ny grams of HCl are consumed in the reaction of \(425 \mathrm{g}\) of a mixture containing \(35.2 \% \mathrm{MgCO}_{3}\) and \(64.8 \% \mathrm{Mg}(\mathrm{OH})_{2},\) by mass? $$\begin{array}{c} \mathrm{Mg}(\mathrm{OH})_{2}+2 \mathrm{HCl} \longrightarrow \mathrm{MgCl}_{2}+2 \mathrm{H}_{2} \mathrm{O} \\ \mathrm{MgCO}_{3}+2 \mathrm{HCl} \longrightarrow \mathrm{MgCl}_{2}+\mathrm{H}_{2} \mathrm{O}+\mathrm{CO}_{2}(\mathrm{g}) \end{array}$$

Short Answer

Expert verified
The total mass of HCl consumed in the reaction is 475.15 grams.

Step by step solution

01

Calculate the amount of each compound in the mixture

First, calculate the amount of each compound in the mixture by applying the percentage by mass. For \(MgCO_3\), we have \(425g * 0.352 = 149.6g\). For \(Mg(OH)_2\), we get \(425g * 0.648 = 275.4g\).
02

Determine the number of moles of each compound

Next, convert the mass of each compound to the number of moles using the molar mass. The molar mass of \(MgCO_3\) is \(24.3g/mol + 12.01g/mol + 3*16.00g/mol = 84.31g/mol\) and that of \(Mg(OH)_2\) is \(24.3g/mol + 2*(1.01g/mol + 16.00g/mol) = 58.32g/mol\). Therefore, there are \(149.6g / 84.31g/mol = 1.774mol\) of \(MgCO_3\) and \(275.4g / 58.32g/mol = 4.723mol\) of \(Mg(OH)_2\).
03

Apply the stoichiometric relationships

For each mole of \(MgCO_3\) and \(Mg(OH)_2\), two moles of HCl are consumed according to the balanced equation. This implies that \(1.774mol * 2 = 3.548mol\) of HCl react with \(MgCO_3\) and \(4.723mol * 2 = 9.446mol\) of HCl react with \(Mg(OH)_2\).
04

Calculate the total mass of HCl

Finally, convert moles of HCl consumed to mass using molar mass of HCl (\(1.01g/mol + 35.45g/mol = 36.46g/mol\)). Hence, the total mass of HCl consumed is \((3.548mol + 9.446mol) * 36.46g/mol = 475.15g\).

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Most popular questions from this chapter

Write a chemical equation to represent the complete combustion of malonic acid, a compound with \(34.62 \% \mathrm{C}, 3.88 \% \mathrm{H},\) and \(61.50 \% \mathrm{O},\) by mass.

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 \(0.981 \mathrm{g} \mathrm{CO}_{2}(\mathrm{g}) ?\) $$\begin{aligned} \mathrm{CaCO}_{3}(\mathrm{s})+2 \mathrm{HCl}(\mathrm{aq}) & \longrightarrow \mathrm{CaCl}_{2}(\mathrm{aq}) +\mathrm{H}_{2} \mathrm{O}(\mathrm{l})+\mathrm{CO}_{2}(\mathrm{g}) \end{aligned}$$

The reaction of calcium hydride with water can be used to prepare small quantities of hydrogen gas, as is done to fill weather-observation balloons. \(\mathrm{CaH}_{2}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \longrightarrow\) $$ \mathrm{Ca}(\mathrm{OH})_{2}(\mathrm{s})+\mathrm{H}_{2}(\mathrm{g}) \text { (not balanced) } $$ (a) How many grams of \(\mathrm{H}_{2}(\mathrm{g})\) result from the reaction of \(127 \mathrm{g} \mathrm{CaH}_{2}\) with an excess of water? (b) How many grams of water are consumed in the reaction of \(56.2 \mathrm{g} \mathrm{CaH}_{2} ?\) (c) What mass of \(\mathrm{CaH}_{2}(\mathrm{s})\) must react with an excess of water to produce \(8.12 \times 10^{24}\) molecules of \(\mathrm{H}_{2} ?\)

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)? $$\mathrm{C}_{6} \mathrm{H}_{11} \mathrm{OH} \longrightarrow \mathrm{C}_{6} \mathrm{H}_{10}+\mathrm{H}_{2} \mathrm{O}$$

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