23.903 grams of product is produced
816.56grams of excess reactant was left over
The reaction between copper solid and sulfur is expressed as:
[tex]Cu(s)+S\rightarrow CuS[/tex]Determine the moles of copper
[tex]\begin{gathered} moles\text{ of Cu}=\frac{7.89\times10^{24}}{6.02\times10^{23}} \\ moles\text{ of Cu}=1.31\times10 \\ moles\text{ of Cu}=13.1moles \end{gathered}[/tex]Determine the moles of sulfur
[tex]\begin{gathered} moles\text{ of S}=\frac{8g}{32} \\ moles\text{ of sulfur}=0.25moles \end{gathered}[/tex]Since sulfur has less moles, hence sulfur is the limiting reactant
Calculate the mass of CuS
Mass = moles * molar mass
Mass of CuS = 0.25 * 95.611
Mass of CuS = 23.903 grams
Determine the mass of excess reactant consumed
[tex]\begin{gathered} Mass\text{ of Cu}=moles\times molar\text{ mass} \\ Mass\text{ of Cu}=0.25\times63.546 \\ Mass\text{ of Cu}=15.8865gram\text{ \lparen mass used up\rparen} \end{gathered}[/tex]Mass of Cu started with = 13.1 * 63.546
Mass of Cu started with = 832.4526grams
Mass of excess reactant left over = 832.4526g - 15.8865g
Mass of excess reactant left over = 816.56grams
Therefore 816.56grams of excess reactant was left over
