How many grams of Nh3 are produced from the reaction of 6.14 g of H2

Question

how many grams of Nh₃ are produced from the reaction of 6.14 g of H₂

N₂ + 3H₂ -> 2NH₃

Answer 1

Hey there Janeriao! Thank you for posting this question. First of all, I was wondering if you could identify what chemistry problem type this question falls under? If you answered a stoichiometry problem, you would be correct! Stoichiometry involves using relationships between reactions & products to figure out some quantitative data; in your case, the amount of grams of Ammonia (NH3) produced from the given amount of 6.14g of H2, given the following unbalanced reaction in your question stem: N2 + 3H2 <--> 2NH3

One of the first steps in solving stoichiometry questions is to (1) balance the equation. Equations may be balanced by counting the quantity of each type of atom in the reactant and product chemical species, followed by placing numerical coefficients in front of each species until all the atoms are balanced.

After balancing your equation, it should appear as follows:

N2 + 3H2 <--> 2NH3 which matches the equation you posted. You want to remember the mole ratio between each species in the reaction: 1 (N2) : 3 (H2) : 2 (NH3)

Normally, you could decipher between which of the reactants is the limiting reagant by looking at mole ratios, going from molar mass to moles of each species present. However, your question specifically asks to determine grams of NH3 from grams of H2.

To perform the calculation, you first start with what is given in the question:

6.14g H2 and convert it to moles by dividing molar mass (g/mol) becomes (mol/g) due to it being in the denominator.

As obtained from the periodic table, the molar mass of Hydrogen is ~1.0g/mol, so we have:

6.14gH2 / 2g/mol => ~3.07molH2 after the units of grams (g) cancel out.

Next, you may convert molH2 into molNH3 which is the final species we want to find grams of.

3.07molH2 x ((2molNH3/3molH2), mole ratios for these species obtained from the coefficients of these species given in the originally balanced reaction).

==> 6.14/3 molNH3. Next, we convert this to our final desired unit of grams by multiplying this value by its molar mass as obtained from the periodic table. N is about 14g/mol. H is about 1.0g/mol. The total MM for NH3 is ~17.0g/mol.

6.14/3 molNH3 x 17.0g/mol ==> mol's cancel out leaving us with: (6.14/3) x (17) g NH3 = ~34.79g NH3.