grams of oxygen how much carbon dioxide could be pro urn methane (CH4)\" if 28.6
ID: 559853 • Letter: G
Question
grams of oxygen how much carbon dioxide could be pro urn methane (CH4)" if 28.6 grams of methane reacts with 57.6 5. Wh 32.1 grams of carbon dioxide are actually produced at is the percentage yield of carbon dioxide in reaction described by question 6 if 6. In the reacti many moles of excess reactant are left at the end of the reaction? on described by question 4 which reactant is the limiting reactant and how 7. A 29.1 gram piece of lead absorbed 38.4 calories of heat energy when it was heated from 23 oC to 64 °C. How much heat would be required to raise 0.0291 kg of lead by the same temperature difference? a. 0.0384 cal b. 3.84 cal c. 38.4 cal d. 38.4 kcalExplanation / Answer
The reaction of combustion of methane is
CH4(g)+2O2(g) --->CO2(g)+ 2H2O(g)
1 mole of CH4 reacts with 2 moles of oxygen to produce 1 mole of CO2
Molar ratio of CH4:O2= 1:2
Moles of CH4= mass/molar mass= 28.6/16= 1.79,
Moles of oxygen = 57.6/32 = 1.8
Actual molar ratio of CH4:O2= 1.79:1.8= 1.79/1.79:1.8/1.79= 1: 1
So excess is CH4 and limiting reactant is Oxygen.
2 mole of O2 gives 1 mole of CO2
1.8 moles of O2 gives 1.8/2=0.9 moles of CO2.
Mass of CO2 produced= moles* molar mass =0.9*44= 39.6 gm
Moles of CH4 consumed = 1.8/2=0.9, mass of CH4 consumed = 0.9*16= 14.4
Mass of CH4 remaining = 28.6-14.4= 14.2 gm
moles of CH4 remaining =14.2/16= 0.8875
Actual amount of CO2 formed = 32.1 gm
% yield= 100* actual yield of CO2/ theoretical yield= 100*32.1/39.6= 81%
2.
We need to determine the specific heat of lead with the given information
Heat absorbed by lead= mass of lead* specific heat of lead* rise in temperature
38.4= 29.1* specific heat*(64-23)
Specific heat of lead= 38.4/(29.1*41)= 0.032 cal/gm.deg.c
When 0.0291 kg of lead is used, mass of lead= 0.0291*1000=29.1 gm=38.2 Cal ( C is correct)
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