When solutions of AgNO3 and NaOH react, the balanced molecular equation is

2AgNO3(aq) + 2NaOH(aq) Ag2O(s) + 2NaNO3(aq) + H2O(l)

How much Ag2O is produced when 0.200 g of AgNO3 and 0.200 g of NaOH react?

calcualate the moles of AgNO3 and moles of NaOH you started with.

Examine the balanced equation, those two react on a one to one mole ratio, so which was the lowest number? that is the most you can produce. So the balanced equation says you get half that (in moles).

To find out how much Ag2O is produced, we need to determine the limiting reactant in the reaction. The limiting reactant is the reactant that is completely consumed and determines the maximum amount of product that can be formed.

To determine the limiting reactant, we can use the concept of stoichiometry, which is the relationship between the amounts of reactants and products in a chemical reaction.

First, let's calculate the number of moles of AgNO3 and NaOH using their respective molecular weights.

Molecular weight of AgNO3:
Ag: 107.87 g/mol
N: 14.01 g/mol
O: 16.00 g/mol (3 oxygen atoms)

So, AgNO3 = 107.87 g/mol + 14.01 g/mol + (16.00 g/mol × 3) = 169.88 g/mol

Number of moles of AgNO3 = Mass of AgNO3 / Molecular weight of AgNO3
= 0.200 g / 169.88 g/mol
≈ 0.001178 mol

Molecular weight of NaOH:
Na: 22.99 g/mol
O: 16.00 g/mol (1 oxygen atom)
H: 1.01 g/mol

So, NaOH = 22.99 g/mol + 16.00 g/mol + 1.01 g/mol = 40.00 g/mol

Number of moles of NaOH = Mass of NaOH / Molecular weight of NaOH
= 0.200 g / 40.00 g/mol
= 0.005 mol

Now, let's compare the stoichiometric ratio of AgNO3 and NaOH in the balanced equation.

From the balanced equation:
2 moles of AgNO3 react with 2 moles of NaOH to produce 1 mole of Ag2O.

The stoichiometric ratio of AgNO3 to NaOH is 2:2 or 1:1.

Since the moles of AgNO3 and NaOH are equal (0.001178 mol vs. 0.005 mol), the ratio is already balanced.

Therefore, both AgNO3 and NaOH are consumed in equal amounts, and the amount of Ag2O produced will be equal to the amount of the limiting reactant.

To calculate the amount of Ag2O produced, we need to convert the moles of AgNO3 to grams using its molecular weight.

Mass of Ag2O = Moles of AgNO3 × Molecular weight of Ag2O
= 0.001178 mol × (2 × 107.87 g/mol)
≈ 0.2514 g

Therefore, approximately 0.2514 grams of Ag2O will be produced when 0.200 grams of AgNO3 and 0.200 grams of NaOH react.

To determine the amount of Ag2O produced when 0.200 g of AgNO3 and 0.200 g of NaOH react, we need to use the concept of stoichiometry and the given balanced molecular equation.

In the balanced molecular equation, we see that the ratio between AgNO3 and Ag2O is 2:1. This means that for every 2 moles of AgNO3, we will produce 1 mole of Ag2O.

To find the moles of AgNO3 and NaOH, we need to divide their masses by their respective molar masses.

The molar mass of AgNO3 is:
Ag: (1 x 107.87 g/mol) = 107.87 g/mol
N: (1 x 14.01 g/mol) = 14.01 g/mol
O: (3 x 16.00 g/mol) = 48.00 g/mol

Total molar mass of AgNO3 = 107.87 + 14.01 + 48.00 = 169.88 g/mol

Moles of AgNO3 = 0.200 g / 169.88 g/mol

Similarly, we can find the moles of NaOH.

The molar mass of NaOH is:
Na: (1 x 22.99 g/mol) = 22.99 g/mol
O: (1 x 16.00 g/mol) = 16.00 g/mol
H: (1 x 1.01 g/mol) = 1.01 g/mol

Total molar mass of NaOH = 22.99 + 16.00 + 1.01 = 40.00 g/mol

Moles of NaOH = 0.200 g / 40.00 g/mol

Since the ratio between AgNO3 and Ag2O is 2:1, we can now determine the moles of Ag2O produced.

Moles of Ag2O = (Moles of AgNO3) / 2

Next, we can use the moles of Ag2O to find the mass of Ag2O produced.

Mass of Ag2O = (Moles of Ag2O) x (Molar mass of Ag2O)

The molar mass of Ag2O is:
Ag: (2 x 107.87 g/mol) = 215.74 g/mol
O: (1 x 16.00 g/mol) = 16.00 g/mol

Total molar mass of Ag2O = 215.74 + 16.00 = 231.74 g/mol

Finally, we can substitute the values into the equation to find the mass of Ag2O produced.

Mass of Ag2O = (Moles of Ag2O) x (Molar mass of Ag2O)
= (0.200 g / 169.88 g/mol) / 2 x 231.74 g/mol

Simplifying the expression will give the answer.