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Chapter 4: Problem 91

When a solid mixture of \(\mathrm{MgCO}_{3}\) and \(\mathrm{CaCO}_{3}\) is heated strongly, carbon dioxide gas is given off and a solid mixture of \(\mathrm{Mg} \mathrm{O}\) and \(\mathrm{CaO}\) is obtained. If a \(24.00 \mathrm{g}\) sample of a mixture of \(\mathrm{MgCO}_{3}\) and \(\mathrm{CaCO}_{3}\) produces \(12.00 \mathrm{g} \mathrm{CO}_{2}\) then what is the percentage by mass of \(\mathrm{MgCO}_{3}\) in the original mixture?

Short Answer

Expert verified
The percentage by mass of \( \mathrm{MgCO}_{3} \) in the original mixture is 95.88%.

Step by step solution

01

Identify the balanced reactions

The reactions for the thermal decomposition of \(\mathrm{MgCO}_{3}\) and \(\mathrm{CaCO}_{3}\) are: \[\mathrm{MgCO}_{3} \rightarrow \mathrm{MgO} + \mathrm{CO}_{2}\] and \[\mathrm{CaCO}_{3} \rightarrow \mathrm{CaO} + \mathrm{CO}_{2}\] From these reactions, one can observe that one mole of either magnesium or calcium carbonate yields one mole of carbon dioxide.
02

Calculate the moles of CO2

The molar mass of \(\mathrm{CO}_{2}\) is 44.01 g/mol. So, to calculate the moles of produced \(\mathrm{CO}_{2}\), divide the given mass of \(\mathrm{CO}_{2}\) by its molar mass. That would be \(12.00 g / 44.01 g/mol = 0.273 mol\) of \(\mathrm{CO}_{2}\) produced.
03

Determine possible mass of each carbonate

Given that one mole of either \(\mathrm{MgCO_{3}}\) or \(\mathrm{CaCO_{3}}\) yields one mole of \(\mathrm{CO2}\), possibly all of the \(\mathrm{CO2}\) could have come from either carbonate. The molar mass of \(\mathrm{MgCO_{3}}\) is 84.33 g/mol and that of \(\mathrm{CaCO_{3}}\) is 100.09 g/mol. So the possible masses of \(\mathrm{MgCO_{3}}\) and \(\mathrm{CaCO_{3}}\) that could have yielded 0.273 mol of \(\mathrm{CO2}\) are \(0.273 mol * 84.33 g/mol = 23.01 g\) and \(0.273 mol * 100.09 g/mol = 27.32 g\) respectively.
04

Calculate the percentage of each

We see that the possible mass of \(\mathrm{MgCO_{3}}\) is less than the total mass of the original sample (which is 24.00 g), while that of \(\mathrm{CaCO_{3}}\) exceeds the total mass. Therefore, the actual mix consists only of \(\mathrm{MgCO_{3}}\) and some additional residue (not contributing to \(\mathrm{CO2}\) production) whose mass can be calculated as: \(24.00 g - 23.01 g = 0.99 g\). Finally, the percentage by mass of MgCO3 in the original mixture is given by: \(23.01 g / 24.00 g * 100% = 95.88% \)

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Most popular questions from this chapter

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