Question

Silver Sulfate reacts with Potassium Chloride according to the following reaction:Ag2SO4 + 2KCl -> 2AgC1 + K2SO4a. If 30.0 grams of Ag2SO4 reacts with 10.0 grams of KCl, what mass of AgCl is produced by the reaction b. the limiting reactant is ______c. how many grams ok K2SO4 can be produced.d. how many grams of excess reactant remain after the reaction e. what is the percent yield if there is 10.0g of AgCl

Answer 1

Explanations:

Given the chemical reaction

`Ag_2SO_4+2KCl\rightarrow2AgCl+K_2SO_4`

Given the following

Mass of Ag2SO4 = 30grams

Mass of KCl = 10grams

Determine the moles of the reactants

`\begin{gathered} mole\text{ of Ag}_2SO_4=\frac{mass}{molar\text{ mass}}molar\text{ mass} \\ mole\text{ of Ag}_2SO_4=(30g)/(311.799) \\ mole\text{ of Ag}_2SO_4=0.0962moles \end{gathered}`
`\begin{gathered} mole\text{ of KCl}=(10g)/(74.5513) \\ moleof\text{ KCl}=0.1341moles \\ 1mole\text{ of KCl}=(0.1341)/(2)=0.06707moles \end{gathered}`

B) B) Since the 1 moles of KCl is lower than the moles of Ag2SO4, hence KCl willl be the limiting reactant.

A) A) According to stoichiometry, 2 moles of KCl produces 2 moles of AgCl, the mass of AgCl produced will be given as;

`\begin{gathered} mass=mole* molar\text{ mass} \\ mass\text{ of AgCl}=0.1341*143.32 \\ mass\text{ of AgCl}=19.22grams \end{gathered}`

C) According to stoichiometry, 2 moles of KCl produces 1 moles of K2SO4, the mole of K2SO4 produced is;

`\begin{gathered} mole\text{ of }K_2SO_4=(1)/(2)*0.1341 \\ mole\text{ of }K_2SO_4=0.06707moles \end{gathered}`
`\begin{gathered} mass\text{ of K}_2SO_4=mole* molar\text{ mass} \\ mass\text{ of K}_2SO_4=0.06707*174.259 \\ mass\text{ of K}_2SO_4=11.69grams \end{gathered}`