Find the limiting reactant first. Then mass of product of NH3. Step by step and where did you get the numbers from.

Question

The equation is 3 H2(g) + N2(g) → 2 NH3(g) First, if _2.00_ g of H2 is reacted with _28.0 g of N2,

Step by step clear explanation on how to get each answer

Answer 1

The first thing we will want to do is convert all of the grams that we are given into moles. We get these numbers from the periodic table of elements.

Molar mass of relevant elements per period table:

Hydrogen - 1.008

Nitrogen - 14.007

Next we can determine the molar mass of each molecule in our chemical equation.

H₂ has 2 H atoms, so the molar mass of this molecule is 1.008 x 2 = 2.016 grams per mole

N₂ has 2 N atoms, so the molar mass of this molecule is 14.007 x 2 = 28.014 grams per mole

NH₃ has 1 N atom and 3 H atoms, so the molar mass of this molecule is (14.007 x 1) + (1.008 x 3) = 17.031 grams per mole

Next we can determine the number of moles of each molecule we have be converting grams (given in the question) to moles using the molar masses we just set up.

To find moles of H₂, we simply divide the number of grams of H₂ (g) by the molar mass of H₂ (grams per mole or g/mol).

2g H₂ ÷ 2.016 g/mol = 0.992 mol of H₂

Do the same for N₂

28g N₂ ÷ 28.014 g/mol = 0.9995 mol of N₂

For this problem, you can say that there is approximately one mole of both H₂ and N₂.

Next we must look at the equation to see which is the rate limiting reactant.

3 H₂(g) + N₂(g) → 2 NH₃(g)

Based on this information, we can see that for every 3 moles of H₂ and 1 mole of N₂, we can produce 2 moles of NH₃.

We should be able to tell that H₂ is the rate limiting reactant by now.

To confirm this, and to determine how much product we will produce, we divide the number of moles that are present, by the number required in the equation for each molecule. The lowest number indicates the rate limiting reactant. For this problem, we have 1 mole of N₂, and our equation requires 1, so 1/1 = 1. Compare this to H₂ of which we have 1 mole, and 3 required by our equation 1/3 = 0.33333. This is the smallest number and is rate limiting. This also indicates how much product we created.

We only made 33.3333% of the product that would have been made if all 3 H₂ atoms were present. This is equal to 0.33333 x 2 moles of NH₃. We only managed to make 0.666667 moles of NH₃. To find the mass product, we simply convert this figure from moles to grams. We can do this using the molar mass we calculated for NH₃ earlier in this problem: 17.031 grams/mol.

This time, rather than dividing to convert from grams to moles, we are multiplying to convert from moles to grams.

0.666667 mol NH₃ x 17.031 grams/mol = 11.354 grams of NH₃ which is the mass of our product.

Answer: H₂ is the rate limiting reactant and the mass of our product is 11.354 grams

Answer 2

In order to solve this problem, we will use the ratios given by the balanced chemical reaction:

3 H₂ + N₂ --> 2 NH₃

Based on this, we require 3 moles of hydrogen gas for every mole of nitrogen gas. We are given 2.00g of Hydrogen gas in the problem along with 28.0g of nitrogen gas.

Since the molar mass of hydrogen gass is 2g/mol, we have 1 mole of hydrogen gas. Since the molar mass of nitrogen gas is 28.0g/mol, we also have one mole of nitrogen gas.

Since we need 3 times the amount of hydrogen based on the balanced equation, we would need 3 moles of hydrogen per 1 mole of nitrogen. Since we do not have 3 moles of hydrogen (we have only 1 mole), Hydrogen gas is the limiting reactant. Thus, we use it to find the total mass of NH₃ formed in the reaction.

3 moles of hydrogen become 2 moles of NH₃ based on the equation.

1 mole H₂ * (2 mol NH₃)/(3 mol H₂) = 2/3 mol NH₃

We then convert this to mass of NH₃ by multiplying by the molar mass of NH₃, which is 14 + 3(1) = 17g/mol

So our final answer is 2/3 mol NH₃ * 17g/mol = 11.33g NH₃