Posted by Dewey on Saturday, April 16, 2011 at 8:50pm.
I find the limiting reagent by solving two simple stoichiometry problems. I use one reagent and find the product, then the other reagent and find the product. Convert 2.3 mol C2H4 to moles of the product.
2.3 mols C2H4 x (1 mole C2H4F2/1 mole C2H2) = 2.3 x (1/1) = 2.3 mols C2H4F2.
Then do the same for HF.
12 moles HF x (1 mole C2H4F2/2 moles HF) = 12 x (1/2) = 6 moles C2H4F2
Obviously both answers can't be right; the correct one in limiting reagent problems is ALWAYS the smaller one. In this case, the limiting reagent is C2H2 and there will be 2.3 moles C2H4F2 formed.
Related Questions
Science/Chemistry - Write a Balanced equation for the chemcial reaction ...
chemistry - Acetylene gas, c2h2, burns in oxygen to produce carbon dioxide and ...
chemistry - The equation for burning C2H2 is 2C2H2(g) + 5O2(g) ⎯&#...
chemistry - what mass of acetylene, C2H2, will be produced from the reaction of ...
Chem B - What mass of acetylene, C2H2, will be produced for the reaction of 90.g...
chemistry - If 10 g of C2H2 is mixed with 80 g of Br2, what masses of c2h2br2 ...
Chemistry 3A - C2H2(g)+ 5/2o2(g)-> 2Co2(g)+ H2o(g)+ 1255.8KJ how many ...
Chemistry - If 10.00 g of C2H2 is mixed with 80.0 g of Br2, what masses of ...
Chemistry - Use the data given to calculate the value of delta G rxn for the ...
Chemistry - The combustion reaction of acetylene gas is represented by this ...
For Further Reading