How many grams of NH3 form when 27.5L of H2 (measured at STP)react with N2 to form NH3 according to:

N2(g) + 3H2 (g) = 2NH3 (g)

Well, to find out how many grams of NH3 form, we'll need to do some calculations.

But before we do that, let me ask you this: why did the scarecrow win an award?

Because he was outstanding in his field! Now, let's get back to the question.

According to the balanced equation, 1 mole of N2 reacts with 3 moles of H2 to form 2 moles of NH3.

Since we measured the volume of H2 at STP, we can use the molar volume of a gas at STP, which is 22.4L/mol.

So, 27.5L of H2 is equal to 27.5/22.4 = 1.23 moles of H2.

Now, let's use the mole ratio from the balanced equation. We have 3 moles of H2 reacting with 2 moles of NH3, so:

(1.23 mol H2) * (2 mol NH3 / 3 mol H2) = 0.82 moles of NH3.

To convert moles of NH3 to grams, we need to use the molar mass of NH3, which is 17.03 g/mol.

So, 0.82 moles of NH3 is equal to 0.82 * 17.03 = 13.96 grams of NH3.

Therefore, approximately 14 grams of NH3 would form when 27.5L of H2 reacts with N2.

Now, isn't it amazing how chemistry can lead to such explosive jokes?

To find the number of grams of NH3 formed, we can use the stoichiometry of the balanced chemical equation.

From the balanced chemical equation:

1 mole of N2 reacts with 3 moles of H2 to produce 2 moles of NH3.

First, let's calculate the number of moles of H2:

Using the Ideal Gas Law equation:

PV = nRT

where:
P = pressure (STP is 1 atm)
V = volume (27.5 L)
n = number of moles (unknown)
R = gas constant (0.0821 L·atm/mol·K)
T = temperature (STP is 273.15 K)

Rearranging the equation to solve for n:

n = PV / RT

Substituting the values:

n = (1 atm) * (27.5 L) / (0.0821 L·atm/mol·K * 273.15 K)
n ≈ 1.06 moles of H2

From the balanced chemical equation, we know that for every 3 moles of H2, 2 moles of NH3 are formed.

So, the number of moles of NH3 formed can be calculated as follows:

moles of NH3 = (2/3) * moles of H2
moles of NH3 = (2/3) * 1.06 moles
moles of NH3 ≈ 0.71 moles of NH3

Finally, we can convert moles of NH3 to grams using the molar mass of NH3.

The molar mass of NH3 is:
(1 atom of N * atomic mass of N) + (3 atoms of H * atomic mass of H) = (1 * 14.01 g/mol) + (3 * 1.01 g/mol) ≈ 17.03 g/mol

Therefore, the number of grams of NH3 formed is:

grams of NH3 = moles of NH3 * molar mass of NH3
grams of NH3 ≈ 0.71 moles * 17.03 g/mol
grams of NH3 ≈ 12.07 grams

So, approximately 12.07 grams of NH3 will be formed when 27.5 L of H2 reacts with N2 under these conditions.

To find the number of grams of NH3 formed, we will need to use stoichiometry. Stoichiometry is a branch of chemistry that deals with the quantitative relationships between reactants and products in chemical reactions.

1. First, let's determine the number of moles of H2 we have. We are given the volume of H2 at STP (Standard Temperature and Pressure). At STP, 1 mole of any gas occupies a volume of 22.4 liters. Therefore, we can calculate the number of moles of H2 as follows:

Moles of H2 = Volume of H2 (in liters) / 22.4 L/mol

Moles of H2 = 27.5 L / 22.4 L/mol

Moles of H2 ≈ 1.23 moles

2. According to the balanced equation, the stoichiometric ratio between H2 and NH3 is 3:2. This means that for every 3 moles of H2, 2 moles of NH3 are formed. Therefore, we can determine the number of moles of NH3 formed by using this ratio:

Moles of NH3 = (Moles of H2 / 3) * 2

Moles of NH3 = (1.23 moles / 3) * 2

Moles of NH3 ≈ 0.82 moles

3. Finally, to find the mass of NH3 formed, we need to multiply the number of moles by its molar mass. The molar mass of NH3 is approximately 17.03 g/mol.

Mass of NH3 = Moles of NH3 * Molar mass of NH3

Mass of NH3 = 0.82 moles * 17.03 g/mol

Mass of NH3 ≈ 13.94 grams

Therefore, approximately 13.94 grams of NH3 will be formed when 27.5 liters of H2 (measured at STP) react with N2 according to the given balanced equation.

Here is an example of a stoichiometry problem. Just follow the steps. To convert L H2 to moles, remember that 1 mole of a gas occupies 22.4 L at STP.

http://www.jiskha.com/science/chemistry/stoichiometry.html

Post your work if you get stuck.