7. Nitric acid (NO) reacts with oxygen gas to form nitrogen dioxide (NO 2 ), a d
ID: 827594 • Letter: 7
Question
7. Nitric acid (NO) reacts with oxygen gas to form nitrogen dioxide (NO2), a dark brown gas:
2NO(g) + O2(g) ? 2NO2(g)
In one experiment 0.886 mole of NO is mixed with 0.503 mol of O2.
a. Calculate which of the two reactants is the limiting reagent.
b. Calculate also the number of moles of NO2 produced.
c. What reactant is left over and how much of it is left over?
8. Propane (C3H8) is a component of natural gas and is used in domestic cooking and heating. It burns according to the following reaction:
C3H8 + O2 ? CO2 + H2O
a. Balance the equation representing the combustion of propane in air.
C3H8 + 5O2 ? 3CO2 + 4H2O ( THIS IS THE ANSWER I CAME UP WITH )
b. How many grams of carbon dioxide can be produced by burning 20.0 pounds of propane, the typical size of a BBG grill propane tank? Assume that oxygen is the excess reagent in this reaction.
Explanation / Answer
a)We see that for every 2 moles of NO gas, we need 1 mole of O2 gas; this yields 2 moles of NO2 gas. If NO was our limiting reagent at 0.886 mol, we would need 0.443 mol of O2. Since the experiment said we had 0.503 mol, then we see that NO is in fact our limiting reagent (simply because it is possible).
b) If we used 0.886 mol of NO, then we would create 0.886 mol of NO2 because it is a 2:2 ratio in the reaction given.
c) Since NO is our limiting reagent, then O2 gas is the leftover reagent. If we use 0.886 mol of NO gas, then we also use 0.443 mol of O2.....so we subtract this from what we started with: 0.503 - 0.443 = 0.06 mol leftover
8a) C3H8 + 5 O2 = 3 CO2 + 4 H2O is the balanced equation
8b) 20 lbs = 9071.85 grams. We need to convert this to moles. 9071.85 g / 44 g = 206.18 mol
Our molar ratio of propane and CO2 is 1:3 ratio. So if we use 206.18 mol of C3H8, then we will produce three times as much CO2 gas. 3*206.18 =618.53 mol of CO2. Convert it to grams by multiplying by 44(formula weight of CO2)
27,215.55 grams of CO2 produced
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