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.
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.
To calculate the grams of solid product, Ag2CO3, in both scenarios, we need to use stoichiometry.
Step 1: Convert the given volume of the solution (10 mL) to moles.
Given:
- Volume of solution = 10 mL
To calculate the moles of the solute (AgNO3 or Na2CO3), we need to know the concentration of the solutions. Let's assume that both solutions are at a concentration of 1 mol/L.
Step 2: Calculate the moles of AgNO3 and Na2CO3 using the given volume and concentration.
For AgNO3:
- Volume = 10 mL = 0.01 L (since 1 L = 1000 mL)
- Concentration = 1 mol/L
Moles of AgNO3 = Volume × Concentration
= 0.01 L × 1 mol/L
= 0.01 mol
For Na2CO3:
- Volume = 10 mL = 0.01 L (since 1 L = 1000 mL)
- Concentration = 1 mol/L
Moles of Na2CO3 = Volume × Concentration
= 0.01 L × 1 mol/L
= 0.01 mol
Step 3: Use stoichiometry to calculate the grams of Ag2CO3 produced.
From the balanced equation:
1 mole of AgNO3 reacts with 1 mole of Ag2CO3
For the first scenario:
Moles of Ag2CO3 produced = Moles of AgNO3 used = 0.01 mol
To convert moles to grams, we need to use the molar mass of Ag2CO3. The molar mass of Ag2CO3 is approximately 275.75 g/mol.
Mass of Ag2CO3 = Moles of Ag2CO3 × Molar mass of Ag2CO3
= 0.01 mol × 275.75 g/mol
Calculate the final result to get the grams of Ag2CO3.
For the second scenario, the steps are the same, but we use the moles of Na2CO3 instead of AgNO3 to calculate the grams of Ag2CO3.
Hope this helps!
To calculate the grams of solid product, Ag2CO3, in Part 1:
Step 1: Determine the molar mass of AgNO3
AgNO3 molar mass = (Ag molar mass) + (N molar mass) + (3 * O molar mass)
= (107.87 g/mol) + (14.01 g/mol) + (3 * 16.00 g/mol)
= 169.87 g/mol
Step 2: Calculate the moles of AgNO3 in 10 mL of the solution
Moles of AgNO3 = (volume in L) * (concentration in mol/L)
= (10 mL / 1000 mL/L) * (34 g / 169.87 g/mol)
= 0.002 mol
Step 3: According to the balanced equation, the mole ratio between AgNO3 and Ag2CO3 is 2:1.
Therefore, the moles of Ag2CO3 formed will be half of the moles of AgNO3 used.
Moles of Ag2CO3 = (0.002 mol) / 2
= 0.001 mol
Step 4: Calculate the grams of Ag2CO3 formed
Grams of Ag2CO3 = (moles of Ag2CO3) * (molar mass of Ag2CO3)
= (0.001 mol) * (273.74 g/mol)
= 0.274 g
Therefore, the grams of solid product, Ag2CO3, from the reaction with 10 mL of AgNO3 containing 34 grams of AgNO3, is approximately 0.274 grams.
-----------------------------------------------------------------------
To calculate the grams of solid product, Ag2CO3, in Part 2:
Step 1: Determine the molar mass of Na2CO3
Na2CO3 molar mass = (2 * Na molar mass) + (C molar mass) + (3 * O molar mass)
= (2 * 22.99 g/mol) + 12.01 g/mol + (3 * 16.00 g/mol)
= 105.99 g/mol
Step 2: Calculate the moles of Na2CO3 in 10 mL of the solution
Moles of Na2CO3 = (volume in L) * (concentration in mol/L)
= (10 mL / 1000 mL/L) * (21 g / 105.99 g/mol)
= 0.001 mol
Step 3: According to the balanced equation, the mole ratio between Na2CO3 and Ag2CO3 is 1:1.
Therefore, the moles of Ag2CO3 formed will be equal to the moles of Na2CO3 used.
Moles of Ag2CO3 = 0.001 mol
Step 4: Calculate the grams of Ag2CO3 formed
Grams of Ag2CO3 = (moles of Ag2CO3) * (molar mass of Ag2CO3)
= (0.001 mol) * (273.74 g/mol)
= 0.274 g
Therefore, the grams of solid product, Ag2CO3, expected from the reaction with 10 mL of Na2CO3 containing 21 grams of Na2CO3, is approximately 0.274 grams.