Answer:
The mass of zinc sulfide produced is [tex]M_{ZnS} = 0.76 \ g[/tex]
Explanation:
From the question we are told that
The mass of zinc is [tex]m_z = 0.750 \ g[/tex]
The mass of sulfur is [tex]m_s = 0.250 \ g[/tex]
The molar mass of [tex]Zn_{(s)}[/tex] is a constant with value 65.39 g /mol
The molar mass of [tex]S_{(s)}[/tex] is a constant with value 32.01 g/mol
The molar mass of [tex]ZnS_{(s)}[/tex] is a constant with value 97.46 g/mol
The reaction is
[tex]Zn_{(s)} + S_{(s)} ------> ZnS_{(s)}[/tex]
So from the reaction
1 mole of [tex]Zn_{(s)}[/tex] react with 1 mole of [tex]S_{(s)}[/tex] to produce 1 mole of [tex]ZnS_{(s)}[/tex]
This implies that
65.39 g /mol of [tex]Zn_{(s)}[/tex] react with 32.01 g/mol of [tex]S_{(s)}[/tex] to produce 97.46 g/mol of [tex]ZnS_{(s)}[/tex]
From the values given we can deduce that the limiting reactant is sulfur cause of the smaller mass
So
0.250 g of [tex]Zn_{(s)}[/tex] react with 0.250 of [tex]S_{(s)}[/tex] to produce [tex]x \ g[/tex] of [tex]ZnS_{(s)}[/tex]
So
[tex]x = \frac{97.46 * 0.250}{32.01}[/tex]
[tex]x = 0.76 \ g[/tex]
Thus the mass of the mass of zinc sulfide produced is
[tex]M_{ZnS} = 0.76 \ g[/tex]
