Given the reaction N2 + 3H2 --> 2NH3...?

Question

How many grams of ammonia are produced when 1.0 mole of nitrogen reacts?

The choices for this are
a.) 8.5 b.)17 c.)34 d.)68

Answer 1

C is correct.

The given (balanced) equation tells you that for 1 mole of N2 you will produce 2 moles of NH3.

N = 14
H = 1(x3) = 3
total mass of NH3 = 17 g / mol

2 mol x 17 g/mol = 34 g

Answer 2

1 mole of nitrogen (N2) makes 2 moles of ammonia. The formual weight of ammonia is the sum of the atomic weight of all the atoms in the molecule

= 1 x N and 3 x H
=14 + 1 +1 +1
=17 g/mol

But each mole of nitrogen makes 2 moles of ammonia, so 1 mol of nitrogen will produce 2 x 17 g of ammonia = 34 g.

Answer 3

look at the balanced equation. For every one mole of N2 there are 2 moles of NH3. Since there is only 1 mol of N2 then there are 2 moles of Nh3. Then use the molar mass of NH3 (17g/mol) and multiply that times 2 to get 34.

Answer 4

I believe you need to use the Equilibrium Constant for this reaction to calculate the number of mols of NH3. Then use the molar mass of NH3 to convert that number of mols to grams. Since you have 1 mol of N2 to start, I think you can assume 3 mol of H2 (since that is the coefficient). Remember in the Keq equation to use 4x^2 for the unknown NH3 (due to it being 2 mols squared).

Answer 5

Go from moles N2-->moles NH3 --> grams NH3
1 mole of N2 * 2 moles of NH3 * 17 grams of NH3 = 34

c.) 34

Answer 6

N2+3H2--->2NH3
1 mole of N2 reacts.
2 moles of NH3 are produced.
NH3 weighs 17.
2 moles =34g
answer c

Answer 7

1.0 moles N(2) yields 2.0 moles of NH(3), by the reaction above.

NH(3) has 17 grams/mole.

17.0 [g/mol] * 2.0 [mol NH(3)] = 34.0 grams NH(3)

Answer 8

enthapy of formation of ammonia is enthalpy per 1 molecule of ammonia, and since enthalpy of formation of 2 molecules is 92kJ, enthapy for 1 molecule is 92kJ/2 = 46kJ

Answer 9

c.) 34