Notes
The stoichiometry path
- Always start with a balanced equation. Coefficients give mole ratios.
- Grams A $\rightarrow$ (÷ molar mass A) $\rightarrow$ moles A $\rightarrow$ (× mole ratio) $\rightarrow$ moles B $\rightarrow$ (× molar mass B) $\rightarrow$ grams B.
- For gases at STP, use 22.4 L/mol; at the same temperature and pressure, volume ratio = mole ratio.
Limiting reactant and yield
- Find the product each reactant could make. The one that makes less is the limiting reactant; the other is in excess.
- Theoretical yield comes from the limiting reactant.
- $\displaystyle \text{percent yield} = \frac{\text{actual}}{\text{theoretical}} \times 100\%$.
Energy
| Exothermic | Endothermic | |
|---|---|---|
| Heat | released | absorbed |
| Sign of $\Delta H$ | negative | positive |
| Surroundings | get warmer | get colder |
- Scale $\Delta H$ by moles reacted. Heat for water: $q = mc\Delta T$, $c = 4.18$ J/(g·°C).
Topics in this unit
Each topic has its own short notes and practice.
Practice
$\displaystyle 6.0 \times \frac{2}{3} = 4.0$ mol.
$\displaystyle \frac{4.86}{24.31} = 0.200$ mol Mg $\rightarrow$ 0.200 mol MgO, and $0.200 \times 40.31 = 8.06$ g.
Al needs $\displaystyle 4.0 \times \frac{3}{2} = 6.0$ mol $\text{Cl}_2$, but only 4.0 mol is present, so $\text{Cl}_2$ runs out first.
$\displaystyle \frac{10.0}{12.5} \times 100\% = 80.0\%$.
The $-572$ kJ is for 2 mol $\text{H}_2$, so 1.00 mol releases $\displaystyle \frac{572}{2} = 286$ kJ.
Heat flows from your hand into the pack, so the reaction absorbs energy.
Free response
Write your answer first, then open the scoring guide.
Free-response question
A student places 1.35 g of aluminum foil into a solution containing 20.0 g of copper(II) chloride. The reaction is $2\text{Al} + 3\text{CuCl}_2 \rightarrow 2\text{AlCl}_3 + 3\text{Cu}$. (Al = 26.98, Cu = 63.55, Cl = 35.45)
(a) Calculate the moles of Al and the moles of $\text{CuCl}_2$.
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Al: $\displaystyle \frac{1.35}{26.98} = 0.0500$ mol. $\text{CuCl}_2$ molar mass $= 63.55 + 2(35.45) = 134.45$ g/mol, so $\displaystyle \frac{20.0}{134.45} = 0.149$ mol.
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(b) Identify the limiting reactant and find the theoretical yield of copper in grams.
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To use all the Al: $\displaystyle 0.0500 \times \frac{3}{2} = 0.0750$ mol $\text{CuCl}_2$ needed. 0.149 mol is available, so Al is limiting. Cu formed $\displaystyle = 0.0500 \times \frac{3}{2} = 0.0750$ mol, and $0.0750 \times 63.55 = 4.77$ g.
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(c) The student dries the copper and finds a mass of 4.29 g. Calculate the percent yield and give one reason it is below 100%.
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Percent yield $\displaystyle = \frac{4.29}{4.77} \times 100\% = 89.9\%$. Possible reasons: some copper stuck to the beaker or passed through the filter paper, or the reaction did not go to completion.
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