5.1g of NH3 and 3g of NO are mixed and kept in a container of 1L at 27°c.find total pressure exerted by gases
Answers
Answer:
In other words, the total pressure is the sum of the individual partial pressures.And Dalton's Law of partial pressures
Law of partial pressures assures us that in a gaseous mixture, the partial pressure exerted by a component gas is the same as the pressure it would exert if it ALONE occupied the container.And we can use the Ideal Gas equation to estimate the pressure:
Law of partial pressures assures us that in a gaseous mixture, the partial pressure exerted by a component gas is the same as the pressure it would exert if it ALONE occupied the container.And we can use the Ideal Gas equation to estimate the pressure:P=nR TV
VSo all we have to do is to solve for P gas
gas individually:
gas individually:P
H
2
2=
2=1.00
2=1.00⋅
2=1.00⋅g
2=1.00⋅g2.016
2=1.00⋅g2.016⋅
2=1.00⋅g2.016⋅g
2=1.00⋅g2.016⋅g⋅
2=1.00⋅g2.016⋅g⋅m
2=1.00⋅g2.016⋅g⋅mo
2=1.00⋅g2.016⋅g⋅mol
2=1.00⋅g2.016⋅g⋅mol−
2=1.00⋅g2.016⋅g⋅mol−1
2=1.00⋅g2.016⋅g⋅mol−1×
2=1.00⋅g2.016⋅g⋅mol−1×0.0821
2=1.00⋅g2.016⋅g⋅mol−1×0.0821⋅
2=1.00⋅g2.016⋅g⋅mol−1×0.0821⋅L
2=1.00⋅g2.016⋅g⋅mol−1×0.0821⋅L⋅
2=1.00⋅g2.016⋅g⋅mol−1×0.0821⋅L⋅a
2=1.00⋅g2.016⋅g⋅mol−1×0.0821⋅L⋅at
2=1.00⋅g2.016⋅g⋅mol−1×0.0821⋅L⋅atm
2=1.00⋅g2.016⋅g⋅mol−1×0.0821⋅L⋅atmK
2=1.00⋅g2.016⋅g⋅mol−1×0.0821⋅L⋅atmK⋅
2=1.00⋅g2.016⋅g⋅mol−1×0.0821⋅L⋅atmK⋅m
2=1.00⋅g2.016⋅g⋅mol−1×0.0821⋅L⋅atmK⋅mo
2=1.00⋅g2.016⋅g⋅mol−1×0.0821⋅L⋅atmK⋅mol
2=1.00⋅g2.016⋅g⋅mol−1×0.0821⋅L⋅atmK⋅mol×
2=1.00⋅g2.016⋅g⋅mol−1×0.0821⋅L⋅atmK⋅mol×300
2=1.00⋅g2.016⋅g⋅mol−1×0.0821⋅L⋅atmK⋅mol×300⋅
2=1.00⋅g2.016⋅g⋅mol−1×0.0821⋅L⋅atmK⋅mol×300⋅K
2=1.00⋅g2.016⋅g⋅mol−1×0.0821⋅L⋅atmK⋅mol×300⋅K1.00
2=1.00⋅g2.016⋅g⋅mol−1×0.0821⋅L⋅atmK⋅mol×300⋅K1.00
Given info : 5.1g of NH3 and 3g of NO are mixed and kept in a container of 1L at 27° C.
To find : The total pressure exerted by gases is ...
solution : given weight of NH3 = 5.1 g
molecular weight of NH3 = 17g
so, mole of NH3 = 5.1/17 = 0.3
similarly, given weight of NO = 3g
molecular weight of NO = 30g
so, mole of NO = 3/30 = 0.1
chemical reaction between NH3 and NO is given by, 4NH3(g) + 6NO(g) ⇒5N2(g) + 6H2O (l)
here 4 mol of NH3 reacts with 6 mol of NO
so, 0.3 mol of NH3 reacts with 6/4 × 0.3 = 0.45 mol of NO.
but given only 0.1 mol
so, NO is limiting reagent.
Now 6 mol of NO produces 5 mol of N2
so, 0.1 mol of NO produces 5/6 × 0.1 = 0.083 mol of N2
also 0.1 mol of NO reacts with 4/6 × 0.1 = 0.067 mol of NH3
total mol of gases remaining in the vessel = 0.083 mol + (0.3 - 0.067) mol
= 0.316 mol
now using PV = nRT
here, V = 1L , n = 0.316 mol, R = 0.082 atm.l/mol.k , T = 300K
⇒P × 1 = 0.316 × 0.082 × 300
⇒P = 7.77 atm
Therefore the pressure exerted by gases is 7.77 atm.