Question
Asked Oct 22, 2019

If 125 mL of nitrogen gas, measured at 756 torr and 23.5 oC, react with excess iodine according to the following reaction, which mass of nitrogen triiodide is produced? 

N2 (g)+ 3I2(s) --> 2NI3(s)

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Step 1

To calculate the mass of nitrogen triiodide produced first it is required to calculate the number of moles of nitrogen, which can be done by using the following equation,

PV nRT
P 756 torr
1 torr 0.001315 atm
756 torr
0.9947 atm
V = 125 mL
0.125 L
T
23.5 °C 273 23.5 296.5 K
R 0.0821 L atm mol-1 K-1
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PV nRT P 756 torr 1 torr 0.001315 atm 756 torr 0.9947 atm V = 125 mL 0.125 L T 23.5 °C 273 23.5 296.5 K R 0.0821 L atm mol-1 K-1

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Step 2

Substituting the values to calculate moles of nitrogen,

PV nRT
0.9947 x 0.125 nx 0.0821 x 296.5
n 0.0051 moles
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PV nRT 0.9947 x 0.125 nx 0.0821 x 296.5 n 0.0051 moles

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Step 3

Now as per the balanced chemical reaction, 1 mole of nitrogen forms 2 moles of nitrogen triiodide,

Hence 0.0051 moles will f...

Mass
No. of moles
Molecular Mass
Molecular mass of NI3= 14 3 x 127
395 g/mol
Mass
No. of moles
Molecular Mass
Mass
0.0102
395
Mass = 4.03 g
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Mass No. of moles Molecular Mass Molecular mass of NI3= 14 3 x 127 395 g/mol Mass No. of moles Molecular Mass Mass 0.0102 395 Mass = 4.03 g

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