Stoichiometry help?

Question

Could someone please help me with this problem, as I do not understand it. Here it is,

2NH4NO3 --------> 2N2 + 4H2O + O2

What is the total number of liters of gas formed when 228 G NH4NO3 is decomposed? (Assume STP.)

Answer 1

You need to find the total no of moles of gas. 1 mole of NH4NO3 will produce 1 mole of N2 and 1/2 mole of O2; the molecular mass of NH4NO3 is 80 g/mole, so 228g is 228/80 or 2.85 moles. Then there will be 2.85 moles of N2 and 1.425 moles of O2, or a total of 4.275 moles of gas. At STP, the volume of gas is 22.4 L/mole, so there will be 95.76 L of gas

Answer 2

so, for O2:
okay so you want to convert gàmoles, and in order to do so, you need to know the molar mass of NH4NO3, which is 2(14) + 4(1) + 3(16) = 80 g/mol
so
(228 g NH4NO3/1) x (1 mol NH4NO3/ 80 g NH4NO3)

and you want to get volume of O2(g) so you use mole ratio, which is 1mol O2/2 mol NH4NO3

so (228 g NH4NO3/1) x (1 mol NH4NO3/ 80 g NH4NO3) x (1 mol O2/2 mol NH4NO3)

and since you’re assuming it’s at STP (as in question), you know that at STP, for 1 mol, there is 22.4 L for volume

so, multiply by 22.4

(228 g NH4NO3/1) x (1 mol NH4NO3/ 80 g NH4NO3) x (1 mol O2/2 mol NH4NO3) x (22.4 L O2/1mol O2) = 31.92 L O2
and since you need sig figs, and you’re given 228 g, so there’s 3 sig figs
so
31.9 L O2

and for N2:
you do the same thing. But instead of mole ratio: 1 mol O2/2 mol NH4NO3, it’s 2 mol N2/2 mol NH4NO3 and
so:
(228 g NH4NO3/1) x (1 mol NH4NO3/ 80 g NH4NO3) x (2 mol N2/2 mol NH4NO3) x (22.4 L N2/1mol N2) = 63.84 L N2 or 63.8 L N2

and for the total, just add the two up so:
31.92 + 63.84 = 95.76 L gas or 95.8 L gas is your answer

Answer 3

First, let me restate your question.

2 NH4NO3 --> 2 N2 + 4 H2O + O2
= 2 moles of NH4NO yields 2 moles of N2, 4 moles of H2O and 1 mole of O2

Now, we use stoichiometry, with the help of ratio and proportions, to solve this problem.

GIVEN:
228 g NH4NO3

ASKED:
L N2 + L H2O + L O2

-------------

First, we convert NH4NO3 to moles.
We use the molar mass (g/mol) of the compound.

N = 14.01
H = 1.01
O = 16.00

N2 = 2(14.01)
H4 = 4(1.01)
O3 = 3(16)
NH4NO3 = 80.06 g/mol

mol NH4NO3 = (228 g) / (80.06 g/mol)
= 2.85 mol

-------

Now, use ratio and proportion.

NH4NO3 : N2
2 : 2 = 2.85 mol NH4NO3 : 2.85 mol N2

NH4NO3 : O2
2 : 1 = 2.85 mol NH4NO3 : 1.425 mol H2O

** Since this deals with nitric acid, H2O is a liquid and thus we do not have to consider.

Σ mol = 2.85 + 1.425
= 4.275 mol

Σ L = (22.4 L/mol)(4.275 mol)
= 95.76 L

[[ STP --> standard temp n pressure; 22.4 L/mol for any gas ]]

: = Therefore, 95.76 L of gas is formed when NH4NO3 is decomposed.

Answer 4

2 moles of NH4NO3 will yield 3 moles of gases (2 moles of N2 and 1 mole of O2; since I am to assume STP, 4 moles H2O will be a liquid, not a gas).
228 g NH4NO3 = 2.848 moles (228 g / 80.0 g/mole)
2.848 moles x (3/2) = 4.273 moles of gas
PV=nRT
1 * V = 4.273 * 0.08206 * 273
V = 95.8 liters

Answer 5

what percentage moles of nitrogen and hydrogen are required to make one mole of ammonia, NH3? equation N2 + 3H2 = 2NH3 so dividing via 2 to get one mole of NH3 provides a million/2 mole of N2 and 3/2 mole of H2