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.
