Help Please this is my last problem!!!?

Question

99

When a mixture of 10.0g of acetylene (C_2H_2) and 10.0g of oxygen(O_2) is ignited, the resultant combustion reaction produces CO_2 and H_2O.

(a) write the balanced chemical equation for this reaction.

(b)which is the limiting reactant?

(c) how many grams of C_2H_2, O_2 , CO_2, and H_2O are present after the reaction is complete?

Please show all work and explain if u want

I have listed this question several times to get in noticed so please just answer one of them

thanks for ur help and yes i tried on my own first and got some where but i know its all wrong.

Answer 1

a) write the balanced chemical equation for this reaction.

2 C2H2 & 5 O2 -->  4CO2  & 2H2O

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masses used in the reaction:

2 C2H2 = (2 mles @ 26.0373 g/mol) = 52.07 grams of C2H2

5 O2  = (5moles  @ 18.02g/mole)  = 90.1 grams of O2

4CO2  = (4 moles @ 44.01 g/mol) = 176.04 grams of CO2

2H2O = (2 moles @ 18.01 g/mol) = 36.02 grams of H2O

we notice that it takes more grams of oxygen to react with acetylene, & that they failled to add that extra oxygen,...

your answer is: Oxygen is your limiting reagent

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after the reaction is complete, how many grams are present ....of :

C_2H_2,

? g C2H2 react with 10.0 g O2 @ 52.07 g C2H2 / 90.1 g O2 = 5.78 g C2H2 react,....... 10.0 g @ start  - 5.78 g now =

your answer: 4.2 grams of C2H2 are left over

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O2 ,  none

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CO2,

? g CO2 porduced from 10.0 g O2 @ 176.04 g CO2 / 90.1 g O2 = 19.54 g CO2

your answer: 19.5 grams of CO2 can be made

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H2O

? g CO2 porduced from 10.0 g O2 @ 36.02 g H2O / 90.1 g O2 = 4.00 g H2)

your answer: 4.00 grams of H2O can be made