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Leni [432]
4 years ago
8

Ammonium nitrite undergoes decomposition to produce only gases as shown below. nh4no2(s) → n2(g) + 2h2o(g) how many liters of ga

s will be produced by the decomposition of 32.0 g of nh4no2 at 525°c and 1.5 atm?
Chemistry
1 answer:
schepotkina [342]4 years ago
3 0
From the given balanced reaction equation:

we can see that every 1 mole of NH4NO2 will produce 1 mole of N2 and when the water also on the gas phase so, every 1 mole of NH4NO2 will produce 3 moles of gas

now we need to get number of moles of NH4NO2  = mass/molar mass
when the mass = 32 g and the molar mass of NH4NO2 = 64 mol/g

∴moles of NH4NO2 = 32 g / 64 mol/g = 0.5 mole

so, the moles that produced of gas will be = 3 * 0.5 = 1.5 moles

now, we can use the ideal gas formula to get V:

PV = nRT

when P is the pressure = 1.5 atm

n is the number of moles = 1.5 moles 

R is the ideal gas constant = 0.0821 L*atm/mol*K

and T is the temperature in Kelvin = 525 °C + 273 = 798 K

so by substitution:

1.5 atm * V = 1.5 moles * 0.0821 * 798 K

∴ V = 65.5 L
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2 years ago
What volume of ammonia gas, measured at 547.9 mmHg and 27.6oC, is required to produce 8.98 g of ammonium sulfate according to th
Studentka2010 [4]

Answer:

4.65 L of NH₃ is required for the reaction

Explanation:

2NH₃(g)  +  H₂SO₄(aq)  → (NH₄)₂SO₄(s)

We determine the ammonium sulfate's moles that have been formed.

8.98 g . 1mol / 132.06 g = 0.068 moles

Now, we propose this rule of three:

1 mol of ammonium sulfate can be produced by 2 moles of ammonia

Therefore, 0.068 moles of salt were produced by (0.068 . 29) / 1 = 0.136 moles of NH₃. We apply the Ideal Gases Law, to determine the volume.

Firstly we do unit's conversions:

27.6°C +273 =  300.6 K

547.9 mmHg . 1 atm / 760 mmHg = 0.721 atm

V = ( n . R . T ) / P → (0.136 mol . 0.082 L.atm/mol.K . 300.6K) / 0.721 atm

V = 4.65 L

4 0
3 years ago
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8 0
3 years ago
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MA_775_DIABLO [31]
This uses the concept of freezing point depression. When faced with this issue, we use the following equation:
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Because the freezing point depression is a colligative property, it does not depend on the identity of the molecules, just the number of them.
Now, we know that molality will be constant, and Kf will be constant, so our only unknown is "i", or the van't hoff factor.
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Therefore, AlBr3 will lower the freezing point of water the most.
6 0
3 years ago
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