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Tatiana [17]
4 years ago
6

Given the reaction has a percent yield of 86.8 how many grams of aluminum iodide would be required to yield an actual amount of

73.75 grams of aluminum?
Chemistry
1 answer:
ioda4 years ago
4 0

Answer:

Approximately 1.29 \times 10^3 grams.

Explanation:

Let x represent the number of grams of aluminum iodide required to yield that 73.75 grams of aluminum.  

In most cases, the charge on each aluminum ion would be +3 while the charge on each iodide ion would be -1. For the charges to balance, there needs to be three iodide ions for every aluminum ion. Hence, the empirical formula for aluminum iodide would be \rm AlI_3.

How many moles of formula units in that x grams of \rm AlI_3? Start by calculating its formula mass M(\mathrm{AlI_3}). Look up the relative atomic mass of aluminum and iodine on a modern periodic table:

  • Al: 26.982.
  • I: 126.904.

M(\mathrm{AlI_3}) = 1\times 26.982 + 3\times 126.904 = 410.694\; \rm g \cdot mol^{-1}.

n(\mathrm{AlI_3}) = \displaystyle \frac{m}{M} = \frac{x}{410.694}\;\rm mol.

Since there's one aluminum ion in every formula unit,

n(\mathrm{Al}) = n(\mathrm{AlI_3}) = \displaystyle \frac{x}{410.694}\; \rm mol.

How many grams of aluminum would that be?

m(\mathrm{Al}) = n \cdot M = \displaystyle \frac{x}{410.694}\; \times 26.982 = \frac{26.982}{410.694}\, x\; \rm g.

However, since according to the question, the percentage yield (of aluminum) is only 86.8\%. Hence, the actual yield of aluminum would be:

\begin{aligned}&\text{Actual Yield} \\ &= \text{Percentage Yield} \times \text{Theoretical Yield} \\ &= 86.8\% \times \frac{26.982}{410.694}\, x \\ &= 0.868 \times \frac{26.982}{410.694}\, x \\ &\approx 0.0570263\, x\; \rm g\end{aligned}.

Given that the actual yield is 73.75 grams,

0.0570263\, x = 73.75.

\displaystyle x = \frac{73.75}{0.0570263} \approx 1.29 \times 10^3\; \rm g.

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0.216 moles of gas can the container hold if a sealed container can hold 0.325 L of gas at 1.00 atm and 293 K.

<h3>What is an ideal gas equation?</h3>

The ideal gas law (PV = nRT) relates the macroscopic properties of ideal gases. An ideal gas is a gas in which the particles (a) do not attract or repel one another and (b) take up no space (have no volume).

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Given data:

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Putting value in the given equation:

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Hence, 0.216 moles of gas can the container hold if a sealed container can hold 0.325 L of gas at 1.00 atm and 293 K.

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