When a mixture of 12.5g of acetylene (C2H2) and 12.5g of oxygen (O2) is ignited,

Question

When a mixture of 12.5g of acetylene (C2H2) and 12.5g of oxygen (O2) is ignited, the resultant combustion reaction produces CO2 and H2O

A. Write the balanced chemical equation for this reaction?
B. Which is the limiting reactant? Express your answer as a chemical formula.
C. How many grams of C2H2 are present after the reaction is complete?
D. How many grams of O2 are present after the reaction is complete?
E. How many grams of CO2 are present after the reaction is complete?
F. How many grams of H2Oare present after the reaction is complete?

I'm really confused.

Explanation / Answer

balanced equation : 2C2H2 + 5O2 ----> 4CO2+ 2H2O relative molecular masses: C2H2=26.036, O2=32, CO2=44.01, H2O=18.016 Since the reaction occur according to the limiting reagent, we will calculate with the mole of limiting reagent. 12.5g O2 x 1/32g per mole = 0.3906 mole of O2 will be reacting A) 0.3906mole O2 x 2mole C2H2/5mole O2 x 26.036g C2H2 = 4.068g of C2H2 will be reacted. So 12.5g(initial) - 4.068g(reacted) = 8.432g of C2H2 will still be present after the reaction. B) Since O2 is the limiting, no O2 will left after the reaction. All of O2 will be reacted. So,it's 0g. C) 0.3906mole O2 x 4moleCO2/5mole O2 x 44.01g CO2 = 13.75g of CO2 will be formed. D) 0.3906mole O2 x 2moleH2O/5mole O2 x

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