Please Help! Mastering Chemistry A heliox deep-sea diving mixture contains 2.0 g
ID: 967513 • Letter: P
Question
Please Help! Mastering Chemistry
A heliox deep-sea diving mixture contains 2.0 g of oxygen to every 98.0 g of helium.
Part A: What is the partial pressure of oxygen when this mixture is delivered at a total pressure of 7.7 atm? Express your answer using two significant figures.
Ammonium nitrate decomposes explosively upon heating according to the following balanced equation: 2NH4NO3(s)2N2(g)+O2(g)+4H2O(g).
Part A: Calculate the total volume of gas (at 129 C and 768 mmHg ) produced by the complete decomposition of 1.43 kg of ammonium nitrate.
Explanation / Answer
solution: a] find moles of each. Then add them to get total moles.
2/16=0.125 moles of oxygen
98/4=24.5 moles of helium
total moles= 0.125 + 24.5 =24.625
pressureO2= molesO2/totalmoles * 7.7atm = (0.125/24.625) X 7.7= 0.039
b ] 2NH4NO3(s)2N2(g)+O2(g)+4H2O(g)
Molecular wt. of ammonium nitrate = 80.043g
1.43Kg= 1430 g
(1430/80.043)= 17.86 moles of ammonium nitrate
from the balance eqaution for every two moles of ammonium nitrate that react, 2 moles of N2, 4 moles of water, one mole of oxygen are produced. that means for every 2 moles of ammonium nitrate 7 moles of gases are produced.
therefore 17.86 moles X 7/2= 62.51 moles of gases are formed.
by using formula PV=nRT
V=nRT/P = 62.51mol X 0.08206 x L atm mol-1 K-1(129+ 273.16)K/ 1atm
= 2062.91L