When 1 mol of sodium nitrate, is dissolved in 1 cubic decimeter of water, 40 kJ of heat energy is absorbed. What is the drop in
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Answer:
percent yield = 40.6 %
Explanation:
The question asks to determine the percent yield, which can be defined as:
percent yield =
where the actual yield is how much product was obtained, in this case 12.5 g of CCl₂F₂, and the theoretical yield is how much product could be obtained with the given reactants theoretically.
So we know already the actual yield, we need to <em>calculate the theoretical yield.</em>
First we need to <em>write the reaction chemical equation</em>:
CCl₄ + HF → CCl₂F₂ + HCl
and <em>balance the equation</em>:
CCl₄ + 2 HF → CCl₂F₂ + 2 HCl
In the equation we can see that <em>for every mol of CCl₄ we should get 1 mol of CCl₂F₂</em> (molar ratio 1:1). So if we <u>calculate the moles of CCl₄</u> in the given 39.2 g of CCl₄ we could know how many moles of CCl₂F₂ (assuming HF is in excess).
- Moles of CCl₄ = mass CCl₄ / molar mass CCl₄
- Molar Mass CCl₄ = 12.011 + 4 * 35.45 = 153.811 g/mol
- Moles of CCl₄ = 39.2 g / 153.811 g/mol = 0.2549 moles
From the molar ratio we know:
Moles of CCl₂F₂ = moles of CCl₄ = 0.2549 moles
Now we need to <u>convert these moles into grams</u> to get the theoretical yield of CCl₂F₂ in grams:
- mass CCl₂F₂ = moles CCl₂F₂ * molar mass CCl₂F₂
- Molar Mass CCl₂F₂ = 12.011 + 2 * 35.45 + 2 * 18.998 = 120.907 g/mol
- Mass CCl₂F₂ = 0.2549 moles * 120.907 g/mol = 30.81 g
- Theoretical yield CCl₂F₂ = 30.81 g
Percent yield = (12.5 g / 30.81 g) * 100 = 40.6 %