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First of all you have to determine the equation for the reaction: C3H8 + 5 O2 -----> 3CO2 + 4H2O Now we determine the limiting reactant: C3H8= (12*3) + (8*1)=44g/mol 8g C3H8 * (1mol/44g ) * (5 O2 / 1 C3H8) * (32g/1mol O2) = 29.09 g O2 So, 8g of C3H8 would use up 29.09 grams of O2 in order for this reaction to go to compleation. Since we have only 8 g of O2, O2 will run out first. So, it is considered the limiting reactant. In the next step we use the same method to determine mass of CO2 formed: 8g O2 * (1mol O2 /32g) * (3CO2 / 5 O2) * (44g/1mol CO2) = 6.6 g CO2 if you have any more questions feel free to e-mail me on micuninikola@yahoo.com I ll be glad to help out Nick Nikolic, Chemistry 121 student

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Q: How many grams of CO2 are used when 8.0g of O2 are produced?
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