Question

Can you please show me the steps of how to get the excess reactant

Answer 1

23.903 grams of product is produced

816.56grams of excess reactant was left over

Explanations:

The reaction between copper solid and sulfur is expressed as:

`Cu(s)+S\rightarrow CuS`

Determine the moles of copper

`\begin{gathered} moles\text{ of Cu}=(7.89*10^(24))/(6.02*10^(23)) \\ moles\text{ of Cu}=1.31*10 \\ moles\text{ of Cu}=13.1moles \end{gathered}`

Determine the moles of sulfur

`\begin{gathered} moles\text{ of S}=(8g)/(32) \\ moles\text{ of sulfur}=0.25moles \end{gathered}`

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

`\begin{gathered} Mass\text{ of Cu}=moles* molar\text{ mass} \\ Mass\text{ of Cu}=0.25*63.546 \\ Mass\text{ of Cu}=15.8865gram\text{ \lparen mass used up\rparen} \end{gathered}`

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