Question

# NH3(g) + O2(g) --> NO2(g) + H2O(g) Consider the above unbalanced equation. What volume of NH3...

NH3(g) + O2(g) --> NO2(g) + H2O(g) Consider the above unbalanced equation. What volume of NH3 at 870 mm Hg and 17.3°C is needed to react with 180 mL of O2 at 724 mm Hg and 21.8°C? Please provide the answer in mL

4 NH3 + 7 O2 ---> 4 NO2 + 6 H2O

step 1: calculate mol of O2

we have:

P = 724.0 mm Hg

= (724.0/760) atm

= 0.9526 atm

V = 180.0 mL

= (180.0/1000) L

= 0.18 L

T = 21.8 oC

= (21.8+273) K

= 294.8 K

find number of moles using:

P * V = n*R*T

0.9526 atm * 0.18 L = n * 0.08206 atm.L/mol.K * 294.8 K

n = 7.085*10^-3 mol

step 2: find mol of NH3

from reaction:

mol of NH3 = (4/7)*moles of O2

= (4/7)*7.085*10^-3 mol

= 4.049*10^-3 mol

step 3: find volume

we have:

P = 870.0 mm Hg

= (870.0/760) atm

= 1.1447 atm

T = 17.3 oC

= (17.3+273) K

= 290.3 K

we have below equation to be used:

P * V = n*R*T

1.1447 atm * V = 4.049*10^-3 mol* 0.0821 atm.L/mol.K * 290.3 K

V = 0.0843 L

V = 84.3 mL

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