For the following reaction, 0.620 grams of hydrogen gas are allowed to react with 12.7 grams of ethylene (c2h4). hydrogen (g) + ethylene (c2h4) (g) ethane (c2h6) (g) what is the maximum amount of ethane (c2h6) that can be formed? grams what is the formula for the limiting reagent? what amount of the excess reagent remains after the reaction is complete? grams

The answer to your question is:

a) 9.3 g of C2H6

b) Hydrogen

c) 3.92 grams of C2H4

Explanation:

Data

H2 = 0.62 g

C2H4 = 12.7 g

MW H2 = 2g

MW C2H4 = 28 g

MW C2H6 = 30 g

Balanced reaction

                                 H2(g) + C2H4 (g)   ⇒     C2H6 (g)

a)        

For H2

                   2 g of H2 1 mol

                   0.62 g       x

                   x = 0.31 mol H2

For C2H4

                  28 g of C2H4  1 mol

                  12.7 g                 x

                   x = 0.45 mol of C2H4

Limiting reactant = H2 because the proportion H2 :: C2H4 is 1 :: 1 and there are more moles of C2H4 than moles of H2.

                    1 mol of H2 1 mol of C2H6

                    0.31 mol         x

                   x = 0.31 mol of C2H6

                    1 mol of C2H6      30 g

                   0.31 mol               x

                    x = 9.3 g of C2H6            

c)

Excess reactant

                         mol of C2H4 = initial moles - H2 moles

                                               =   0.45 - 0.31 mol

                                               = 0.14 mol

                         1 mol of C2H4 28 g

                         0.14 mol              x

                        x = 3.92 grams

it would be b. 6co2 + 6h2o --> c6h12o6 + 6o2