Chemistry: How many moles of gas are present in a 1.07 l sample at 1.56 atm and a temperature of 22.0oc?

How many moles of gas are present in a 1.07 l - id1071017773, alayjared12374

Ответ:

lindseydupre

10.01.2020

Chemistry

Approximately $24\; \rm g$ (at most.)

Explanation:

Aluminum $\rm Al$ reacts with bromine $\rm Br_2$ at a 2:3 ratio:

$\rm 2\; Al\, (s) + 3\; Br_2\, (g) \to 2\; AlBr_3\, (s)$ .

Look up the relative atomic mass of $\rm Al$ and $\rm Br$ . From a modern periodic table:

$\rm Al$ : 26.982

. $\rm Br$ : 79.904

Calculate the formula mass of the reactants and of the product:

$M(\mathrm{Al}) = 26.986\; \rm g \cdot mol^{-1}$ . $M(\mathrm{Br_2}) = 2\times 79.904 = 159.808\; \rm g \cdot mol^{-1}$ . $M(\mathrm{AlBr_3}) = 26.986 + 3 \times 79.904 = 266.698\; \rm g \cdot mol^{-1}$ .

Calculate the quantity (in number of moles of formula units) of each reactant:

$\displaystyle n(\mathrm{Al}) = \frac{m(\mathrm{Al})}{M(\mathrm{Al})} = \frac{5.0\; \rm g}{26.986\; \rm g \cdot mol^{-1}} \approx 0.18528\; \rm mol$ . $\displaystyle n(\mathrm{Br_2}) = \frac{m(\mathrm{Br_2})}{M(\mathrm{Br_2})} = \frac{22\; \rm g}{159.808\; \rm g \cdot mol^{-1}} \approx 0.13767\; \rm mol$ .

Assume that $\rm Al\, (s)$ is the limiting reactant. From the coefficients:

$\displaystyle \frac{n(\mathrm{AlBr_3})}{n(\mathrm{Al})} = 1$ .

Based on the assumption that $\rm Al\, (s)$ is the limiting reactant:

$\begin{aligned}n(\mathrm{AlBr_3}) &= \frac{n(\mathrm{AlBr_3})}{n(\mathrm{Al})} \cdot n(\mathrm{Al}) \\ &=1\times 0.18528\; \rm mol \approx 0.185\; \rm mol\end{aligned}$ .

In other words, if $\rm Al$ is the limiting reactant (meaning that $\rm Br_2$ is in excess,) then approximately $0.556\; \rm mol$ of $\rm AlBr_3$ will be produced.

On the other hand, assume that $\rm Br_2\; (g)$ is the limiting reactant. Similarly, from the coefficients:

$\displaystyle \frac{n(\mathrm{AlBr_3})}{n(\mathrm{Br_2})} = \frac{2}{3}$ .

Based on the assumption that $\rm Br_2\, (g)$ is the limiting reactant:

$\begin{aligned}n(\mathrm{AlBr_3}) &= \frac{n(\mathrm{AlBr_3})}{n(\mathrm{Br_2})} \cdot n(\mathrm{Br_2}) \\ &= \frac{2}{3}\times 0.13767\; \rm mol \approx 0.0918\; \rm mol\end{aligned}$ .

Compare the $n(\mathrm{AlBr_3})$ value based on the two assumptions. Only the smallest value, $n(\mathrm{AlBr_3}) \approx 0.0918\; \rm mol$ (under the assumption that $\rm Br_2\, (g)$ is the limiting reactant,) would resemble the theoretical yield. The reason is that $\rm Br_2\, (g)$ would run out before all that $\rm 5.0\; g$ of $\rm Al\, (s)$ was converted to $\rm AlBr_3\, (g)$ .

Apply the formula mass of $\rm AlBr_3$ to find the mass of that (approximately) $0.0918\; \rm mol$ of $\rm AlBr_3$ formula units:

$\begin{aligned}m(\mathrm{AlBr_3}) &= n(\mathrm{AlBr_3}) \cdot M(\mathrm{AlBr_3}) \\ &= 0.0918\; \rm mol \times 266.698\; g \cdot mol^{-1} \approx 24\; \rm g\end{aligned}$ .

0,0(0 оценок)

How many moles of gas are present in a 1.07 l sample at 1.56 atm and a temperature of 22.0oc?

Ответ:

Ответ:

Ask your question to the AI advisor

More tips

Answers on questions: Chemistry

Ask an AI advisor a question