Question

How many sulfur atoms are there in 21.0 g of al2s3?

Answer 1

Given the mass of aluminum sulfide$Al_{2}S_{3}$ = 21.0 g

Molar mass of $Al_{2}S_{3}$ = $2 * Molar mass of Al + 3 *molar mass of S = 2(27g/mol)+3(32.g/mol)=150.g/mol$

Calculating the moles of $Al_{2}S_{3}$:

$21.0 g *\frac{1 mol}{150 g} = 0.14 mol Al_{2}S_{3}$

Each mole $Al_{2}S_{3}$ has 3 mol S.

Each mole S constitutes $6.022*10^{23} atoms of S$

Calculating atoms of S in 0.14 mol $Al_{2}S_{3}$:

$0.14 mol Al_{2}S_{3} *\frac{3mol S}{1 mol Al_{2}S_{3} }*\frac{6.022*10^{23}atoms S }{1 mol S}$

= $2.53 * 10^{23} atoms of S$