Question

Directions: Write the balanced equation for each of the following situations. . In addition, list the reaction type. YOU MUST TELL THE AMOUNTS OF EVERY SUBSTANCE THAT REMAINS IN THE CONTAINER AT THE END OF THE REACTION. ASSUME THAT ALL REACTIONS GO TO COMPLETION. If only STOICHIOMETRY, tell how much of the excess reactant is used!!!! Reaction Type a. Combination Reaction b. Decomposition Reaction c. Single Displacement / THIS IS ONE TYPE OF Oxidation Reduction Reaction d. Precipitation Reaction e. Gaseous Reaction f. Neutralization Reaction g. Combustion Reaction 2. 61.802 cg of nitrogen gas is reacted with 61.802 cg of hydrogen gas to form ammonia 2. Balanced Chemical Equation Reaction Type: At the completion of reactions: Grams of nitrogen: Grams of hydrogen: Grams of ammonia:

Answer 1

`3H_2+N_2\rightarrow2NH_3`

Reaction type: Combination (or Synthesis) reaction.

We are given:

mass of N2 = 61.802 cg = 0.61802g

mass of H2 = 61.802 cg = 0.61802g

Lets start by calculating the number of moles of N2 and H2.

For N2:

n = m/M where m is the mass and M is the molar mass

n = 0.61802g/28.0134

n = 0.0218 mol

For H2:

n = m/M

n = 0.61802g/2.01588

n = 0.07977 mol

So to know which one is the excess reactant, we have to divide the moles by thier respective co-efficient.

For N2: 0.0218 mol/1 = 0.0218

For H2: 0.07977 mol/3 = 0.0266

Therefore H2 is the excess reagent. Therefore the limiting reagent is N2 so we will use it to calculate the mass of NH3 produced, because the limiting reagent in a chemical reaction is the reactant that will be consumed completely. Once there is no more of that reactant, the reaction cannot proceed.

Masses after the reaction: We will use the stoichiometry to calculate these masses.

The molar ratio between N2 and NH3 is 1:2, therefore number of moles of NH3 is = 0.0218 x 2 = 0.0436 mol

mass of NH3 = n x M

m = 0.0436 mol x 17,031 g/mol

m = 0.743 g of NH3

Initially we had a mass of 0.61082g H2 and 0.61082 g N2 = 1.222g

mass produced is 0.743g which means 1.222 g-0.743g = 0.478g

Since H2 is the excess reactant, after the reaction there is 0.478g of H2

and all of N2 was used up since it is the limiting reagent. N2 = 0g after reaction.