Question #6572b

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Question #6572b

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Answer 1 · 2015-02-06T08:53:08

Start with the balanced chemical equation

$2 A l B r_{3 (a q)} + 3 C l_{2 (g)} \to 2 A l C l_{3 (a g)} + 3 B r_{2 (l)}$ $2AlBr_(3(aq)) + 3Cl_(2(g)) -> 2AlCl_(3(ag)) + 3Br_(2(l))$

Now, take the mole ratio one of the reactants (you can pick either one) has with all the other compounds in the reaction, this will help you determine whether or not you have a limiting reagent and exactly how much of each product is made in the reaction.

Take, for instance, $A l B r_{3}$ $AlBr_3$ . Notice that it has a $2:3$ $"2:3"$ mole ratio with chlorine gas, i.e. for every 2 moles of $A l B r_{3}$ $AlBr_3$ , you need 3 moles of $C l_{2}$ $Cl_2$ . Any other relationship between the number of moles of these two compounds will result in a limiting reagent.

So, the number of moles of $A l B r_{3}$ $AlBr_3$ is

$47.6 g \cdot \frac{1 mole}{266.7 g} = 0.1785 moles$ $"47.6 g" * ("1 mole")/("266.7 g") = "0.1785 moles"$

Even before calculating the actual number, we've established that you would need

$0.1785 moles A l B r_{3} \cdot \frac{3 moles C l_{2}}{2 moles A l B r_{3}} = 0.2678 moles$ $"0.1785 moles" AlBr_3 * ("3 moles" Cl_2)/("2 moles" AlBr_3) = "0.2678 moles"$ $C l_{2}$ $Cl_2$

If you have less than this, chlorine will be the limiting reagent; if you have more, aluminium bromide will be the limiting reagent.

The number of moles of chlorine can be determined using the molar volume of a gas at STP; under these conditions, 1 mole of any ideal gas occupies $22.4 L$ $"22.4 L"$ . SInce you've got more than that, you'll have more than one mole of gas

$40.0 L \cdot \frac{1 mole}{22.4 L} = 1.786 moles$ $"40.0 L" * ("1 mole")/("22.4 L") = "1.786 moles"$ $C l_{2}$ $Cl_2$

You have way more than the required 0.2678 moles of chlorine, which means that the aluminum bromide will act as the limiting reagent. This means that you'll have excess chlorine

$n_{chlorine in excess} = 1.786 - 0.2678 = 1.518 moles$ $n_("chlorine in excess") = "1.786" - "0.2678" = "1.518 moles"$

The volume of chlorine that will still be in the container after the reaction is complete is

$1.518 moles \cdot \frac{22.4 L}{1 mole} = 34.0 L$ $"1.518 moles" * ("22.4 L")/("1 mole") = "34.0 L"$

Now for the products. Once again use mole ratios to determine how many moles are produced

$0.1785 moles A l B r_{3} \cdot \frac{1 mole A l C l_{3}}{1 mole A l B r_{3}} = 0.1785 moles$ $"0.1785 moles" AlBr_3 * ("1 mole" AlCl_3)/("1 mole" AlBr_3) = "0.1785 moles"$ $A l C l_{3}$ $AlCl_3$

The mass produced will be

$0.1785 moles \cdot \frac{133.3 g}{1 mole} = 23.8 g$ $"0.1785 moles" * ("133.3 g")/("1 mole") = "23.8 g"$ $A l C l_{3}$ $AlCl_3$

$0.1785 moles A l B r_{3} \cdot \frac{3 moles B r_{2}}{2 moles A l B r_{3}} = 0.2678 moles$ $"0.1785 moles" AlBr_3 * ("3 moles" Br_2)/("2 moles" AlBr_3) = "0.2678 moles"$ $B r_{2}$ $Br_2$

$0.2678 moles \cdot \frac{159.8 g}{1 mole} = 42.8 g$ $"0.2678 moles" * ("159.8 g")/("1 mole") = "42.8 g"$ $B r_{2}$ $Br_2$

Therefore, the container will have

$34.0 L$ $"34.0 L"$ chlorine gas, $23.8 g$ $"23.8 g"$ $A l C l_{3}$ $AlCl_3$ , and $42.8 g$ $"42.8 g"$ $B r_{2}$ $Br_2$ ,

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Question #6572b

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