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The decomposition of 86.5 g ammonium nitrate yields how many liters of dinitrogen monoxide at 1.28 atm and 328k

NH4NO3->N2O+2H2O

1 Answer

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First, we calculate the moles of ammonium nitrate:
Moles = mass / molecular mass
Moles = 86.5 / 80
Moles = 1.08

From the equation, it is visible that the molar ratio between ammonium nitrate and dinitrogen monoxide is 1 : 1. The moles of dinitrogen monoxide formed will be:
1.08 moles

Next, we may apply the ideal gas law equation to find the volume:
PV = nRT
V = nRT/P
V = (1.08 * 0.082 * 328)/1.28
V = 22.69 liters

22.7 liters of dinitrogen monoxide will be produced
User Grexis
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