Question

Considering the limiting reactant, what is the mass of zinc sulfide produced from 0.250 g of zinc and 0.750 g of sulfur? Zn(s)+S(S) ZnS(s)

Answer 1

Answer:

The mass of zinc sulfide produced is  $M_{ZnS} = 0.76 \ g$

Explanation:

From the question we are told that

   The mass of zinc is  $m_z = 0.750 \ g$

    The mass of sulfur is  $m_s = 0.250 \ g$

The molar mass   of  $Zn_{(s)}$  is a constant with value  65.39 g /mol

The molar mass of $S_{(s)}$  is a constant with value  32.01 g/mol

The molar mass of  $ZnS_{(s)}$ is a constant with value 97.46  g/mol

The reaction is  

        $Zn_{(s)} + S_{(s)} ------> ZnS_{(s)}$

   So from the reaction

       1 mole of  $Zn_{(s)}$ react with 1 mole of  $S_{(s)}$ to produce 1 mole of $ZnS_{(s)}$

This implies that

65.39 g /mol of  $Zn_{(s)}$ react with 32.01 g/mol of  $S_{(s)}$ to produce   97.46  g/mol  of $ZnS_{(s)}$

From the values given we can deduce that the limiting reactant is sulfur cause  of the smaller mass

 So  

    0.250 g of  $Zn_{(s)}$ react with 0.250 of  $S_{(s)}$ to produce $x \ g$ of  $ZnS_{(s)}$

So

      $x = \frac{97.46 * 0.250}{32.01}$

       $x = 0.76 \ g$

Thus the mass of the mass of zinc sulfide produced is

    $M_{ZnS} = 0.76 \ g$