[tex]\text{ Zn}_{(\text{s\rparen}}\text{ + S}_{(\text{s\rparen}}\text{ }\rightarrow\text{ ZnS}_{(\text{s\rparen}}\text{ }[/tex][tex]\text{ 6.00 g Zn }\times\frac{1\text{ mol Zn}}{65.4\text{ g Zn}}=\text{ 0.0917 mol Z}[/tex][tex]3.35\text{ g S }\times\frac{1\text{ mol S}}{32.07\text{ g S}}=0.1045\text{ mol S}[/tex]
Since the reaction is 1:1, the limiting reactant is Zn and the excess reactant is S.
Mass in excess of S
[tex]\text{ mass = 32.07 g/mol * \lparen}\frac{3.35\text{ g}}{32.07\text{ g/mol}}-\frac{6\text{ g}}{65.4\text{ g/mol}}\text{\rparen = 0.4078 g S}[/tex]
So the answer is option b sulfur 0.407 g
