Nitric acid reacts with silver metal:

4HNO3+3Ag yields NO+2H2O+3 AgNO3.
Calculate the number of grams of NO formed when 10.8 g of Ag reacts with 12.6g of HNO3.

This is a limiting reagent (LR) problem. You know that because amounts are given for BOTH reactants.

mols Ag = grams/atomic mass = ?
mols HNO3 = grams/molar mas = ?

Using the coefficients in the balanced equation, convert mols Ag to mols NO formed.
Do the same for mols HNO3 to mols NO.
It is likely that you will have two different values for mols NO which means one of the values is wrong. The correct value in LR problems is ALWAYS the smaller value and the reagent producing that value is the limiting reagent.
Now taker the smaller value for mols and convert to grams. g = mols x molar mass.

To calculate the number of grams of NO formed, we need to determine the limiting reagent first. The limiting reagent is the reactant that is completely consumed in the reaction and limits the amount of product that can be formed.

First, let's calculate the number of moles of Ag and HNO3 using their respective molar masses.

The molar mass of Ag is 107.87 g/mol:
Number of moles of Ag = Mass of Ag / Molar mass of Ag
Number of moles of Ag = 10.8 g / 107.87 g/mol = 0.1 mol

The molar mass of HNO3 is 63.01 g/mol:
Number of moles of HNO3 = Mass of HNO3 / Molar mass of HNO3
Number of moles of HNO3 = 12.6 g / 63.01 g/mol = 0.2 mol

Next, we need to determine the stoichiometry of the reaction between Ag and HNO3. From the balanced equation, we can see that the ratio of Ag to NO is 3:1.

Since the ratio of Ag to NO is 3:1, we can say that the number of moles of NO formed is equal to 0.1 mol of Ag.

Finally, let's calculate the mass of NO:
Mass of NO = Number of moles of NO x Molar mass of NO
The molar mass of NO is 30.01 g/mol.

Mass of NO = 0.1 mol x 30.01 g/mol = 3.001 g

Therefore, the number of grams of NO formed when 10.8 g of Ag reacts with 12.6 g of HNO3 is 3.001 grams.

To calculate the number of grams of NO formed in the reaction, we need to determine the limiting reactant first. The limiting reactant is the reactant that will be completely consumed and determines the amount of product formed.

Let's calculate the number of moles for each reactant using their respective molar masses:

1) Ag (silver):
Molar mass of Ag = 107.87 g/mol
Number of moles of Ag = mass / molar mass = 10.8 g / 107.87 g/mol = 0.1 mol (approximately)

2) HNO3 (nitric acid):
Molar mass of HNO3 = 63.01 g/mol
Number of moles of HNO3 = mass / molar mass = 12.6 g / 63.01 g/mol = 0.2 mol (approximately)

Now, we need to determine the stoichiometric ratio between Ag and NO in the balanced equation. According to the balanced equation:

4HNO3 + 3Ag → NO + 2H2O + 3AgNO3

The ratio between Ag and NO is 3:1. Therefore, for every 3 moles of Ag that react, 1 mole of NO is formed.

Considering the molar ratio, we can calculate the number of moles of NO that can be formed from the limiting reactant (Ag):

Number of moles of NO = (Number of moles of Ag) × (1 mole of NO / 3 moles of Ag)
= 0.1 mol × (1 mol / 3 mol)
= 0.033 mol (approximately)

Finally, we can convert the number of moles of NO to grams using its molar mass:

Molar mass of NO = 30.01 g/mol
Mass of NO = Number of moles × molar mass
= 0.033 mol × 30.01 g/mol
≈ 0.990 g

Therefore, approximately 0.990 grams of NO are formed when 10.8 grams of Ag reacts with 12.6 grams of HNO3.