Dear Student, Please find below the solution to the asked query: The reaction is AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq) First, determine the moles of the reactants taking part in the reaction Mass of AgNO3 present is =16 9100× 50 = 8 45g Moles of AgNO3 present in the reaction mixture = 8 45 g170 g mol-1 = 0 05 mol Mass of NaCl present is 5 8100× 50 = 2 9 g Moles of NaCl present in the reaction mixture = 2 9 g58 5 g mol-1 = 0 05 mol As 1 mole each of the two reactants form 1 mole of solid AgCl, moles of AgCl formed is 0 05 mol Mass of AgCl formed = 0 05 × 143 = 7 15g Hope this information will clear your doubts about stoichiometry If you have any more doubts just ask here on the forum and our experts will try to help you out as soon as possible Regards

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Dear Student,
Please find below the solution to the asked query:

The reaction is AgNO₃(aq) + NaCl(aq)

\to

AgCl(s) + NaNO₃(aq)

First, determine the moles of the reactants taking part in the reaction.

Mass of AgNO₃ present is =

\frac{16.9}{100} \times 50

= 8.45g
Moles of AgNO₃ present in the reaction mixture =

\frac{8.45 g}{170 {g mol}^{- 1}}

= 0.05 mol
Mass of NaCl present is

\frac{5.8}{100} \times 50

= 2.9 g
Moles of NaCl present in the reaction mixture =

\frac{2.9 g}{58.5 {g mol}^{- 1}}

= 0.05 mol
As 1 mole each of the two reactants form 1 mole of solid AgCl, moles of AgCl formed is 0.05 mol
Mass of AgCl formed = 0.05

\times

143 = 7.15g

Hope this information will clear your doubts about stoichiometry.
If you have any more doubts just ask here on the forum and our experts will try to help you out as soon as possible.
Regards

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