Chemistry I questions?
1 mole of oxygen gas and 2 moles of ammonia are placed in a container and allowed to react at 850 degrees Celsius according to the equation: 4NH3(g) + 5O2(g) --> 4NO(g) + 6H2O(g) a) If the total pressure in the container is 5.00 atm, what are the partial pressures for the three gases remaining? b) Using Graham's Law, what is the ratio of the effusion rates of NH3(g) to O2(g)?
2.
2.00 g of hydrogen gas and 19.2 g of oxygen gas are placed in a 100.0 L container. These gases react to form H2O(g). The temperature is 38 degrees Celsius at the end of the reaction. a) What is the pressure at the conclusion of the reaction? b) If the temperature was raised to 77 degrees Celsius, what would the new pressure be in the same container?
3.
Calculate the molecular weight of a gas if 35.44 g of the gas stored in a 7.50 L tank exerts a pressure of 60.0 atm at a constant temperature of 35.5 °C.
4.
0.190 g of a gas occupies 250.0 mL at STP. What is its molar mass? What gas is it?
Answer:
1.
The gases react in the ratio 4:5. So the limiting reagent is oxygen. So, we have all of oxygen reacting and 1.25 moles of NH3 reacting. So there is a left over 0.75 moles of NH3 in the container. This produces 1 mole of NO and 1.5 moles of H2O. So in all there are (1+1.5+0.75) = 2.25 moles of gases in the container. So, the partial pressures are
p[NH3] = 0.75/2.25*5 = 1.667 atm
p[H2O] = 1.5/2.25 * 5 = 3.333 atm
p[NO] = 1/2.25*5 = 2.222 atm
The rates of effusion are inversely proportional to the square roots of the molecular weights. So we have NH3:O2 = squareroot(32/17)=1.371
2.
The reaction is written as
2H2 + O2 -----> 2H2O
Again use the concept of limiting reagent. We find that O2 is the limiting reagent. So, we have all of the oxygen reacting and (19.2/32*2) = 1.2 grams. So we have 0.8 grams of H2 remaining. That is 0.8/2 = 0.4 moles. We have 1.2 moles of water being formed. But this is liquid at this temperature. Use the universal gas equation.
PV = nRT
P = nRT/V = 0.4*0.0821*311/100 = 0.102 atm
Again P is directly proportional to T.
So, at 77 degrees, the pressure is 0.102*(350/311)=0.1147 atm
3.
Again use the universal gas equation.
PV = nRT
n = PV/RT
n = 60*7.5/(0.0821*308.5)
n = 17.7 moles
35.44 grams is 17.7 moles
So 1 mole is 2 grams. The gas in consideration is thus hydrogen (H2)
4.
Each mole of gas occupies 22.4 L or 22400 mL at STP. So, number of moles of gas = 250/22400=1.11*10^-2 moles
This is 0.190 gram. So 1 mole is 17.024 grams
So, this gas is ammonia.
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