M. 15

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Kelly Cai 4D
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Joined: Sat Jul 20, 2019 12:17 am

M. 15

Postby Kelly Cai 4D » Fri Oct 04, 2019 12:53 pm

Aluminum metal reacts with chlorine gas to produce aluminum chloride. In one preparation, 225 g of aluminum is placed in a container holding 535 g of chlorine gas. After reaction ceases, it is found that 300. g of aluminum chloride has been produced.
(a.) Write the balanced equation for the reaction.
(b.) What mass of aluminum chloride can be produced by these reactants?
(c.) What is the percentage yield of aluminum chloride?

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Joined: Tue Sep 24, 2019 12:17 am

Re: M. 15

Postby 005162520 » Fri Oct 04, 2019 1:39 pm

You will first write the chemical equation that is given and balance it.
Al(s) + Cl2(g) ---> AlCl3(s)
After you balance the chemical equation, you find the moles of Al and Cl2 to help you determine the limiting reactant.
To find the moles, you would convert to moles. So for Al, you would use 255g and divide by its molar mass of Al that would you give you the moles of Al. And for Cl2, you would use 300.g and divide by its molar mass. After finding the moles, you would now compare it to the molar ratio which is a 2:3 ration. To find the mass (which is also another way of asking for the theoretical yield), you would then have to use the limiting reactant and multiply it the ratio 2/3 and convert your answer using the molar mass. (c) to find the percent yield, you use the formula given % yield= (actual yield/ theoretical yield).

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