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.
(please show steps please, I need to understand how you got the answer, thanks)
mols Ag (part a) = g/molar mass = about 0.2 but you need to do it more accurately.
Using the coefficients in the balanced equation convert mols AgNO3 to mols Ag2CO3. That is 0.2 mol AgNO3 x (1 mol Ag2CO2/2 mols Ag_) = 0.2 x 1/2 = aboaut 0.1 mol.
g Ag2CO3 formed = mols x molar mass
b. Do the same and solve for mols (then grams) Ag2CO3.
If the question is asking for grams Ag2CO3 with 34g AgNO3 and 21g Na2CO3 it turns into a limiting reagent problem in which case the FEWER mols Ag2CO3 will be formed..
To calculate the grams of solid product Ag2CO3 in both parts of the question, we need to use the molar ratios from the balanced equation.
Part 1:
First, let's calculate the molarity of the AgNO3 solution in Part 1.
Molarity (M) = moles of solute / volume of solution in liters
The molar mass of AgNO3 is: Ag = 107.87 g/mol + N = 14.01 g/mol + 3O = 16.00 g/mol × 3 = 48.00 g/mol
So, the moles of AgNO3 can be calculated as follows:
moles of AgNO3 = (34 g) / (48.00 g/mol) = 0.71 mol
Next, we look at the stoichiometric ratio between Ag2CO3 and AgNO3 in the balanced equation:
2 mole of AgNO3 reacts with 1 mole of Ag2CO3
Using the molar ratio, we can determine the moles of Ag2CO3 produced:
Moles of Ag2CO3 = 0.71 mol AgNO3 × (1 mol Ag2CO3 / 2 mol AgNO3) = 0.36 mol Ag2CO3
Finally, we can calculate the mass of Ag2CO3 using its molar mass:
mass of Ag2CO3 = moles of Ag2CO3 × molar mass of Ag2CO3
= 0.36 mol × (108.87 g/mol) = 38.99 g
Therefore, the expected mass of Ag2CO3 in Part 1 of the experiment is 38.99 grams.
Part 2:
Using the same approach, we can calculate the expected mass of Ag2CO3 in Part 2 of the experiment.
First, find the molarity of the Na2CO3 solution:
Molarity = moles of Na2CO3 / volume of solution in liters
The molar mass of Na2CO3 is: Na = 22.99 g/mol × 2, C = 12.01 g/mol, and O = 16.00 g/mol × 3
So, the moles of Na2CO3 can be calculated as follows:
moles of Na2CO3 = (21 g) / (22.99 g/mol × 2 + 12.01 g/mol + 16.00 g/mol × 3) ≈ 0.20 mol
Using the stoichiometric ratio from the balanced equation:
1 mole of Na2CO3 reacts with 1 mole of Ag2CO3
Moles of Ag2CO3 = 0.20 mol Na2CO3 × (1 mol Ag2CO3 / 1 mol Na2CO3) = 0.20 mol Ag2CO3
Finally, calculate the mass of Ag2CO3:
mass of Ag2CO3 = moles of Ag2CO3 × molar mass of Ag2CO3
= 0.20 mol × (108.87 g/mol) ≈ 21.77 g
Therefore, the expected mass of Ag2CO3 in Part 2 of the experiment is approximately 21.77 grams.
To calculate the grams of solid product, Ag2CO3, in both scenarios, we need to use stoichiometry and the given amounts of AgNO3 and Na2CO3.
Step 1: Calculate the moles of AgNO3
To find the moles of AgNO3, we need to use its molar mass. The molar mass of AgNO3 is calculated by summing the atomic masses of each element in the compound:
AgNO3: (Ag x 2) + (N x 1) + (O x 3) = 108 + 14 + 48 = 170 g/mol
Given that 10 mL of AgNO3 contains 34 grams, we can use its molar mass to find the number of moles:
moles = mass / molar mass
moles of AgNO3 = 34 g / 170 g/mol = 0.2 mol
Step 2: Apply stoichiometry
From the balanced equation, we know that the molar ratio between AgNO3 and Ag2CO3 is 2:1. This means that for every 2 moles of AgNO3, we should get 1 mole of Ag2CO3.
Since we have 0.2 moles of AgNO3, we can use the stoichiometric ratio to calculate the moles of Ag2CO3 formed:
moles of Ag2CO3 = moles of AgNO3 x (1 mole Ag2CO3 / 2 moles AgNO3)
moles of Ag2CO3 = 0.2 mol x (1/2) = 0.1 mol
Step 3: Calculate the mass of Ag2CO3
To find the mass of Ag2CO3, we need to use its molar mass:
Ag2CO3: (Ag x 2) + (C x 1) + (O x 3) = 108 + 12 + 48 = 168 g/mol
mass of Ag2CO3 = moles of Ag2CO3 x molar mass of Ag2CO3
mass of Ag2CO3 = 0.1 mol x 168 g/mol = 16.8 grams
Therefore, the expected mass of solid product, Ag2CO3, in the first scenario is 16.8 grams.
Now let's move on to the second scenario.
Step 1: Calculate the moles of Na2CO3
Following the same steps as above, we can calculate the moles of Na2CO3:
moles of Na2CO3 = mass / molar mass
moles of Na2CO3 = 21 g / (22.99 + (12.01 x 2) + (16 x 3)) = 21 g / 106 g/mol ≈ 0.198 mol
Step 2: Apply stoichiometry
Using the stoichiometric ratio from the balanced equation, we can find the moles of Ag2CO3:
moles of Ag2CO3 = moles of Na2CO3 x (1 mole Ag2CO3 / 2 moles Na2CO3)
moles of Ag2CO3 = 0.198 mol x (1/2) ≈ 0.099 mol
Step 3: Calculate the mass of Ag2CO3
mass of Ag2CO3 = moles of Ag2CO3 x molar mass of Ag2CO3
mass of Ag2CO3 = 0.099 mol x 168 g/mol ≈ 16.632 grams
Therefore, the expected mass of solid product, Ag2CO3, in the second scenario is approximately 16.632 grams.