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
Answer: 84.3 mL
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