Asked by Anonymous
In the formula 2C4H10+13O2=10H2O+8CO2 whats the mass of butane needed to produce 79.7g of carbon dioxide.
Answers
Answered by
Devron
Just back track,
76.7g of CO2*(1 mole of CO2/44.01g of CO2)= moles of CO2
8 moles of CO2=2 moles of C4H10
moles of CO2*(2 moles of C4H10/8 moles of CO2)=moles of C4H10
moles of C4H10 *(58.12g of C4H10/mole of C4H10)= mass of C4H10
*******I did not check the balanced equation, so I am assuming that you did that part correctly.
76.7g of CO2*(1 mole of CO2/44.01g of CO2)= moles of CO2
8 moles of CO2=2 moles of C4H10
moles of CO2*(2 moles of C4H10/8 moles of CO2)=moles of C4H10
moles of C4H10 *(58.12g of C4H10/mole of C4H10)= mass of C4H10
*******I did not check the balanced equation, so I am assuming that you did that part correctly.
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