Part 1:
First convert grams of AgNO₃ to moles:
34 g AgNO₃ * (1 mol AgNO₃ / 169.87 g AgNO₃) = 0.2002 mol AgNO₃
Next, use the stoichiometry of the balanced equation to convert moles of AgNO₃ to moles of Ag₂CO₃:
0.2002 mol AgNO₃ * (1 mol Ag₂CO₃ / 2 mol AgNO₃) = 0.1001 mol Ag₂CO₃
Finally, convert moles of Ag₂CO₃ to grams:
0.1001 mol Ag₂CO₃ * (275.75 g Ag₂CO₃ / 1 mol Ag₂CO₃) = 27.6 g Ag₂CO₃
The grams of solid product, Ag₂CO₃, is 27.6 g.
Part 2:
First convert grams of Na₂CO₃ to moles:
21 g Na₂CO₃ * (1 mol Na₂CO₃ / 105.99 g Na₂CO₃) = 0.1981 mol Na₂CO₃
Next, use the stoichiometry of the balanced equation to convert moles of Na₂CO₃ to moles of Ag₂CO₃:
0.1981 mol Na₂CO₃ * (1 mol Ag₂CO₃ / 1 mol Na₂CO₃) = 0.1981 mol Ag₂CO₃
Finally, convert moles of Ag₂CO₃ to grams:
0.1981 mol Ag₂CO₃ * (275.75 g Ag₂CO₃ / 1 mol Ag₂CO₃) = 54.6 g Ag₂CO₃
The grams of solid product, Ag₂CO₃, expected from this reaction is 54.6 g.
Heres a balanced equation:
Na2CO3(aq)+2AgNO3 (aq) --> 2NaNO3(aq)+Ag2CO3 (s)
From this equation, 10 mL of AgNO3 in this contains 34 grams of AgNO3. Calculate the grams of solid product, Ag2CO3.
Part 2:
10 mL of Na2CO3 in this experiment contains 21 grams of Na2CO3. Calculation that the grams of solid product, Ag2CO3, expected from this reaction.
1 answer