calculate the volume of amonia that is produced at stp when 10.5 grams of N2 reacted with excess H2.

To calculate the volume of ammonia produced, we need to use the balanced chemical equation for the reaction between nitrogen gas (N2) and hydrogen gas (H2) to form ammonia (NH3):

N2 + 3H2 → 2NH3

From the balanced equation, we can see that two moles of ammonia are produced for every 1 mole of nitrogen gas (N2) consumed. Therefore, we need to convert the given mass of nitrogen (10.5 grams) to moles.

First, we need to determine the molar mass of nitrogen (N2) by adding the atomic masses of its constituent elements: nitrogen (N).

The atomic mass of nitrogen (N) is approximately 14.01 g/mol. Since there are two nitrogen atoms in one molecule of N2, the molar mass of N2 would be 2 x 14.01 g/mol = 28.02 g/mol.

Next, we can use the molar mass of N2 to convert the given mass of nitrogen (10.5 grams) to moles:

moles of N2 = 10.5 g / 28.02 g/mol ≈ 0.375 mol

From the balanced equation, we know that 2 moles of NH3 are produced for every 1 mole of N2. Therefore, the number of moles of NH3 produced will be twice the number of moles of N2:

moles of NH3 = 2 x 0.375 mol = 0.75 mol

Since the reaction takes place at STP (Standard Temperature and Pressure), we can use the ideal gas law equation, PV = nRT, to find the volume of ammonia gas generated.

At STP, the temperature (T) is 273.15 K, and the pressure (P) is 1 atm. The ideal gas constant (R) is 0.0821 L.atm/mol.K.

Rearranging the ideal gas law equation to solve for volume (V), we have:

V = nRT / P

Plugging in the known values:

V = (0.75 mol) x (0.0821 L.atm/mol.K) x (273.15 K) / (1 atm)

V ≈ 16.37 L

Therefore, the volume of ammonia produced at STP when 10.5 grams of N2 reacts with excess H2 is approximately 16.37 liters.