How much aluminum oxide and how much carbon are needed to prepare 573 g of aluminum by the balanced chemical reaction (below) if the reaction proceeds to 79.3 % yield ?

The number of moles of aluminum oxide required to give the actual yield is obtained from the given mass of aluminum produced by the reaction. The percentage yield is used to obtain the theoretical yield of the aluminum. The number of moles produced theoretically can be used to calculate the amount of aluminum oxide required in moles and when we multiply the value obtained by the molar mass of aluminum oxide, we obtain the mass of aluminum oxide required.

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codedmog101 codedmog101 · Answer 2 · 2020-03-14T14:40:59+01:00

Answer:

1361.166 g Al2O3 and 240.3 g Carbon.

Explanation:

Okay, let us start by writing out the balanced Chemical equation of the reaction. The Chemical equation of the reaction below shows the synthesis of Aluminum from Aluminum oxide and carbon.

4 Al2O3 + 3 C ---------> 4 Al + 3CO2.

From the reaction above 4 moles of aluminum oxide react with 3 moles of 3 carbon to give 4 moles of aluminum and 3 moles of Carbondioxide.

Step one: find the number of moles of Aluminum.

Number of moles = mass / molar mass.

The mass of Aluminum given from the question= 573 g. Therefore, the number of moles, n = 573 / 27 grams per mole = 21.2 moles.

Step two: Calculate the theoretical yield. So, the number of moles of carbon= 21.2 × (3/4) = 15.9 moles of carbon.

Number of moles of Aluminum oxide = 21.2 × ( 2/4). = 10.6 moles of Al2O4.

Note that we are given that the reaction proceed to 79.3 percent yield, then, we can calculate the actual yield.

So, for Aluminum, we have;

21.2 × (100/ 79.3). = 26.7 moles of Aluminum.

==> For carbon; 26.7 × (3/4) = 20.025 moles.

===> For Al2O3; 26.7 × ( 2/4). = 13.35 moles.

Step three: conversion of the moles to grams.

Using the formula;

Mass = moles × molar mass.

==> Mass of carbon = 20.025 × 12 = 240.3 g.

===> Mass of Al2O3 = 13.35 × 101.96.

= 1361.166 g.