Asked by vikki
If 6.0g of CH4 and 5.0g of O2 are used, what is the mass of CO2 produced?
Answers
Answered by
Mathmate
Step 1:Write the chemical equation of the oxidation:
CH4+2O2=CO2+2H2O
This means that 1 mole of CH4 or 2 moles of O2 will give 1 mole of CO2
Step 2: find limiting reagent
CH4: 6/(12+4*1)=6/16=0.4375 mol
O2: 5/(2*16)=5/32=0.15625 mol
Compare with the equation in step 1, it is evident that O2 is the limiting reagent.
Step 3: find quantity of CO2 produced by the limiting reagent.
From chemical equation, we know that 2 mol of O2 will produce 1 mol of CO2.
By proportion,
0.15625 mol of O2 will produce 0.15625/2 mol=0.78125 of CO2.
Find the mass of 0.78125 mol of CO2, round answer to 2 significant figures.
CH4+2O2=CO2+2H2O
This means that 1 mole of CH4 or 2 moles of O2 will give 1 mole of CO2
Step 2: find limiting reagent
CH4: 6/(12+4*1)=6/16=0.4375 mol
O2: 5/(2*16)=5/32=0.15625 mol
Compare with the equation in step 1, it is evident that O2 is the limiting reagent.
Step 3: find quantity of CO2 produced by the limiting reagent.
From chemical equation, we know that 2 mol of O2 will produce 1 mol of CO2.
By proportion,
0.15625 mol of O2 will produce 0.15625/2 mol=0.78125 of CO2.
Find the mass of 0.78125 mol of CO2, round answer to 2 significant figures.
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