Answer
$c_{Al_{2}O_{3}} $= 45.90 wt%
Work Step by Step
Required:
When kaolinite clay $Al_{2}(Si_{2}O_{5})(OH_{4})$ is heated to a sufficiently high temperature, chemical water is driven off. Compute the composition of the remaining product (in weight percent $Al_{2}O_{3})$.
Solution:
Rewriting the chemical formula for the given into $Al_{2}O_{3}-2SiO_{2}-2H_{2}O$ and removing the chemical water, it simplified into $Al_{2}O_{3}-2SiO_{2}$.
Computing the formula weight for each: $Al_{2}O_{3} =(2)(26.98 g/mol) + (3)(16.00 g.mol) = 101.96 g/mol$ $SiO_{2} = (28.09 g/mol) +(2)(16.00 g/mol) = 60.09 g/mol$
Computing the wt% of $Al_{2}O_{3}$:
$c_{Al_{2}O_{3}} = \frac{101.96 g/mol}{101.96 g/mol + (2)(60.09 g/mol)} \times 100 $= 45.90 wt%
