The acetylene tank contains 35.0 mol C2H2, and the oxygen tank contains 84.0 mol O2. How many moles of CO2 are produced when 35.0 mol C2H2 react completely? How do you get 35.0 mol out of C2H2?
@ganeshie8
he might be able to help you b cuz i have no idea:(sry
first lets make the equation of reaction
C2H2 + O2 ---> CO2 + H2O I believe
2C2H2 + 5O2 -->4CO2 + 2H2O
There is the equation
is that balanced
I would think thats what it shows in the question
you can check that it is balanced by adding up the Carbons on the left side, equals to carbons on the right side, etc
I'm very new to this
ok so now lets answer the question
ok
2C2H2 + 5O2 -->4CO2 + 2H2O The acetylene tank contains 35.0 mol C2H2, and the oxygen tank contains 84.0 mol O2. How many moles of CO2 are produced when 35.0 mol C2H2 react completely? How do you get 35.0 mol out of C2H2? 35 mol C2H2 * 26.04 grams / mol C2H2 , that will give you the grams of C2H2
I am using this calculator http://www.lenntech.com/calculators/molecular/molecular-weight-calculator.htm plug in C2H2
also you have 84 mol O2 * 32 g / mol O2 = 2688 grams O2
and we have 35 mol C2H2 * 26.04 grams / mol C2H2 = 911.4 grams C2H2
now we need to use stoichiometry, which means we can think of the balanced chemical equation as a mole equation 2 moles of C2H2 + 5 moles of O2 -->4 moles of CO2 + 2 moles H2O
this says for every 2 moles of C2H2 there is 4 moles of CO2
and for every 5 moles of O2 is 2 Moles of H20?
yes that too
ok and do you see how we got 911.4 grams C2H2 2688 grams of O2
How do we setup the equation to get our mols of CO2
The acetylene tank contains 35.0 mol C2H2, and the oxygen tank contains 84.0 mol O2. How many moles of CO2 are produced when 35.0 mol C2H2 react completely? How do you get 35.0 mol out of C2H2? 35 moles of C2H2 * 4 mol CO2 / 2 mol C2H2 =
so i multiplied 35 moles by the ratio 4moles CO2 for every 2 moles C2H2
notice that the C2H2 cancels
All units should cancel right? Leaving you with CO2
70.0 mol of CO2
correct
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