Solutions of sodium carbonate and silver nitrate react to form solid silver carbonate and a solution of sodium nitrate. A solution containing 3.10 of sodium carbonate is mixed with one containing 4.43 of silver nitrate.

How many grams of silver nitrate are present after the reaction is complete?

Well, it looks like we've got a chemistry problem on our hands. Or rather, in our beakers. Let's dive in!

To find out how many grams of silver nitrate are present after the reaction is complete, we need to determine the limiting reagent. In other words, we need to see which reactant will run out first and determine the amount of product formed from the limiting reagent.

First, let's find the moles of sodium carbonate and silver nitrate in their respective solutions.

The molar mass of sodium carbonate (Na2CO3) is approximately 106 g/mol, so the number of moles of sodium carbonate is:

3.10 g Na2CO3 * (1 mol Na2CO3 / 106 g Na2CO3) = 0.029 mol Na2CO3

The molar mass of silver nitrate (AgNO3) is approximately 169.87 g/mol, so the number of moles of silver nitrate is:

4.43 g AgNO3 * (1 mol AgNO3 / 169.87 g AgNO3) = 0.026 mol AgNO3

Now, let's look at the balanced equation for the reaction:

2 Na2CO3 + AgNO3 -> Ag2CO3 + 2 NaNO3

From the equation, we can see that 2 moles of sodium carbonate react with 1 mole of silver nitrate to produce 1 mole of silver carbonate.

Comparing the moles of sodium carbonate and silver nitrate, we can see that we have more moles of sodium carbonate than silver nitrate. Therefore, sodium carbonate is in excess and silver nitrate is the limiting reagent.

Since 1 mole of silver nitrate produces 1 mole of silver carbonate, the number of moles of silver carbonate formed will be equal to the number of moles of silver nitrate present. In this case, it is 0.026 mol.

The molar mass of silver nitrate (AgNO3) is approximately 169.87 g/mol, so the mass of silver nitrate present will be:

0.026 mol AgNO3 * (169.87 g AgNO3 / 1 mol AgNO3) = 4.42 g AgNO3

Therefore, after the reaction is complete, you will have approximately 4.42 grams of silver nitrate left.

Hope that clears things up! If you have any more chemistry conundrums or need a good chemistry joke, just ask!

To determine the number of grams of silver nitrate present after the reaction is complete, we need to use the stoichiometry of the balanced chemical equation.

The balanced chemical equation for the reaction between sodium carbonate and silver nitrate is:

Na2CO3 + 2AgNO3 -> Ag2CO3 + 2NaNO3

From the balanced chemical equation, we can see that 1 mole of sodium carbonate reacts with 2 moles of silver nitrate to produce 1 mole of silver carbonate and 2 moles of sodium nitrate.

First, let's calculate the number of moles of sodium carbonate and silver nitrate:

Molar mass of sodium carbonate (Na2CO3): 22.99 g/mol (Na) + 12.01 g/mol (C) + (3 x 16.00 g/mol) (O) = 105.99 g/mol

Number of moles of sodium carbonate = mass of sodium carbonate / molar mass of sodium carbonate
= 3.10 g / 105.99 g/mol
= 0.0292 mol (approx)

Molar mass of silver nitrate (AgNO3): 107.87 g/mol (Ag) + 14.01 g/mol (N) + (3 x 16.00 g/mol) (O) = 169.87 g/mol

Number of moles of silver nitrate = mass of silver nitrate / molar mass of silver nitrate
= 4.43 g / 169.87 g/mol
= 0.0261 mol (approx)

Based on the stoichiometry in the balanced equation, we can see that 2 moles of silver nitrate are needed to react with 1 mole of sodium carbonate.

Since the ratio of silver nitrate to sodium carbonate is 2:1, we can determine the limiting reactant by comparing the mole ratio to the actual number of moles available.

The available moles of silver nitrate is 0.0261 mol, which is less than the required moles of silver nitrate (2 x 0.0292 mol = 0.0584 mol).

Therefore, the limiting reactant is silver nitrate, and the reaction will be completed when all the silver nitrate is consumed.

Converting the moles of silver nitrate to grams:

Mass of silver nitrate = number of moles of silver nitrate x molar mass of silver nitrate
= 0.0261 mol x 169.87 g/mol
= 4.43 g

Therefore, 4.43 grams of silver nitrate are present after the reaction is complete.

To find the grams of silver nitrate present after the reaction is complete, we need to determine which reactant is limiting and which one is in excess.

First, we need to write a balanced chemical equation for the reaction:

2 Na2CO3 + AgNO3 -> Ag2CO3 + 2 NaNO3

From the balanced equation, we can see that the mole ratio between Na2CO3 and AgNO3 is 2:1. This means that for every 2 moles of Na2CO3, we need 1 mole of AgNO3 to complete the reaction.

Step 1: Convert the given masses of Na2CO3 and AgNO3 to moles.
Given mass of Na2CO3 = 3.10 g
Given mass of AgNO3 = 4.43 g

Molar mass of Na2CO3 = 22.99 g/mol (sodium) + 12.01 g/mol (carbon) + 3(16.00 g/mol) (oxygen) = 105.99 g/mol
Molar mass of AgNO3 = 107.87 g/mol (silver) + 14.01 g/mol (nitrogen) + 3(16.00 g/mol) (oxygen) = 169.87 g/mol

Number of moles of Na2CO3 = Given mass / Molar mass = 3.10 g / 105.99 g/mol ≈ 0.0292 mol
Number of moles of AgNO3 = Given mass / Molar mass = 4.43 g / 169.87 g/mol ≈ 0.0261 mol

Step 2: Determine the limiting reactant.
From the balanced chemical equation, we know that we need 2 moles of Na2CO3 for every 1 mole of AgNO3. Therefore, for the reaction to be complete, we need at least 0.0292 mol (2 × 0.0146 mol) of AgNO3.

Since 0.0261 mol of AgNO3 is less than 0.0292 mol, AgNO3 is the limiting reactant.

Step 3: Calculate the grams of AgNO3 remaining.
To determine the grams of AgNO3 remaining, we subtract the moles of AgNO3 reacted (0.0261 mol) from the initial moles of AgNO3.

Grams of AgNO3 remaining = (Initial moles - Reacted moles) × Molar mass
= (0.0292 mol - 0.0261 mol) × 169.87 g/mol
= 0.0031 mol × 169.87 g/mol
≈ 0.526 g

Therefore, after the reaction is complete, there will be approximately 0.526 grams of silver nitrate remaining.

2AgNO3 + Na2CO3 ==> 2NaNO3 + Ag2CO3(s)

Before we can do anything you must identify 3.10 what and 4.43 what.
This is a limiting reagent problem (LR) and you know that because amounts are given for BOTH reactants. I work these the long way. I will assume those are grams.
mols AgNO3 = grams/molar mass = ?
mols Na2CO3 = grams/molar mass = ?

Using the coefficients in the balanced equation, convert mols AgNO3 to mols Ag2CO3.
Do the same and convert mols Na2CO3 to mols Ag2CO3.
It is likely that the two values will not agree which means one of them is not right; the correct value in LR problems is ALWAYS the smaller value and the reagent responsible for producing that number is the LR.

Now use the smaller number of mols Ag2cO3 and convert that to grams Ag2CO3.
g = mols x molar mass.