Question

How many sulfur atoms are present in 25.6 g of Al2(S2O3)3

Answer 1

Given the mass of $Al_{2}(S_{2}O_{3})_{3}$=25.6 g

The molar mass of $Al_{2}(S_{2}O_{3})_{3}$=390.35g/mol

Converting mass of $Al_{2}(S_{2}O_{3})_{3}$to moles:

$25.6 g Al_{2}(S_{2}O_{3})_{3}*\frac{1molAl_{2}(S_{2}O_{3}}{390.35 gAl_{2}(S_{2}O_{3}} =0.0656molAl_{2}(S_{2}O_{3}$

Converting mol $Al_{2}(S_{2}O_{3})_{3}$to mol S:

$0.0656mol Al_{2}(S_{2}O_{3})_{3}*\frac{6molS}{1mol Al_{2}(S_{2}O_{3})_{3}}=0.3936 molS$

Converting mol S to atoms of S using Avogadro's number:

1 mol = $6.022*10^{23}atoms$

$0.3936mol S *\frac{6.022*10^{23}atoms S}{1 mol S}=2.37*10^{23} S atoms$