b) ( I guess amount of Na2O2 = 28 g, image cropped improperly,) work =P x V PV =
ID: 700457 • Letter: B
Question
b) ( I guess amount of Na2O2 = 28 g, image cropped improperly,)
work =P x V
PV = nRT
V = nRT / P
work = P x ( nRT / P)
work = nRT
n = number of moles of gas
2 Na2O2 (s) + 2 H2O (l) -----> 4 NaOH (aq) + O2 (g)
2 moles of Na2O2 can produce 1 moloe of O2
moles of Na2O2 = 28 g / 77.98 g/mol => 0.3590 moles
So moles of O2 produced = 0.3590 moles / 2 => 0.1795 moles
R = 8.314 J/mol.K
T = 30 + 273 = 303 K
work = 0.1795 mol x 8.314 J/mol.K x 303 K
work = 452.3 J
14) 215 kcal = 215000 cal
215000 cal x ( 4.184 J / 1 cal) => 899560 J
or 899.56 kJ
Explanation / Answer
b) Calculate the amount of work done (in Joules) against an atmospheric pressure of 1.05 atm when g of Nad. reacts with excess water at 30.0°C 14 How many joules are equivalent to 215 kilocalories? (1 cal-4.184 )
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