Answer:
The co2 si the e theing for he h2o for the photosysthis of the moon= mass
Explanation:
Answer:
2C2H6(g) + 7O2(g) ==> 4CO2(g) + 6H2O(g)
One way to find the limiting reactant is to divide the moles of reactant by the corresponding coefficient in the balanced equation and whichever is less is limiting.
For C2H6 we have 2 moles / 2 = 1
For O2 we have 8 moles / 7 = 1.14
Explanation:
If a reaction mixture contains 8 moles of O2(g) with 2 moles of C2H6(g),which reactant is the limiting reactqnt and what is the mass (in grams) remaining of the reactant in excess?
C
It is a idea which no kind of evidence is allowed to prove wrong
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