Silver reacts with nitric acid to form nitrogen monoxide, silver nitrate and water. Calculate the volume of nitrogen monoxide gas (@STP) that can be produced from the reaction of 5.00g of silver?

3Ag + 4HNO3 ==> NO + 2H2O + 3AgNO3

mols Ag = grams/atomic mass Ag
Use the coefficients in the balanced equation to convert mols Ag to mols NO.
Then L NO = mols NO x 22.4 L/mol =

0.346 L NO

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To calculate the volume of nitrogen monoxide gas produced, we need to use the balanced chemical equation for the reaction. Here is the balanced equation:

2Ag + 4HNO3 → 2AgNO3 + 2NO + 2H2O

From the equation, we can see that 2 moles of silver (Ag) react with 4 moles of nitric acid (HNO3) to produce 2 moles of nitrogen monoxide (NO).

1. Convert the mass of silver (Ag) to moles:
To convert grams to moles, we need to divide the given mass by the molar mass of silver (Ag). The molar mass of silver is 107.87 g/mol.

Number of moles of Ag = Given mass of Ag / Molar mass of Ag

Given mass of Ag = 5.00 g
Molar mass of Ag = 107.87 g/mol

Number of moles of Ag = 5.00 g / 107.87 g/mol = 0.0463 mol

2. Use the mole ratio to find the moles of nitrogen monoxide (NO):
From the balanced equation, we know that 2 moles of Ag react to produce 2 moles of NO. Therefore, the moles of NO produced will be equal to the moles of Ag.

Number of moles of NO = Number of moles of Ag = 0.0463 mol

3. Convert the moles of nitrogen monoxide (NO) to volume (@STP):
At standard temperature and pressure (STP), one mole of any ideal gas occupies 22.4 L of volume.

Volume of NO gas (@STP) = Number of moles of NO x Molar volume (@STP)

Number of moles of NO = 0.0463 mol
Molar volume (@STP) = 22.4 L/mol

Volume of NO gas (@STP) = 0.0463 mol x 22.4 L/mol = 1.039 L

Therefore, the volume of nitrogen monoxide gas (@STP) that can be produced from the reaction of 5.00 g of silver is approximately 1.039 liters.