Answer:
A. 10.29 g of C2H6.
B. C2H4.
C. 3.14 g of H2.
Explanation:
Step 1:
We'll begin by writing the balanced equation for the reaction. This is given below:
C2H4 + H2 —> C2H6
Step 2:
Determination of the masses of C2H4 and H2 that reacted and the mass of C2H6 produced from the balanced equation.
This is illustrated below:
Molar mass of C2H4 = (12x2) + (4x1) =28 g/mol
Mass of C2H4 from the balanced equation = 1 x 28 = 28 g.
Molar mass of H2 = 2x1 = 2 g/mol
Mass of H2 from the balanced equation = 1 x 2 = 2 g.
Molar mass of C2H6 = (12x2) + (6x1) = 30 g/mol.
Mass of C2H6 from the balanced equation = 1 x 30 = 30 g
From the balanced equation above,
28 g of C2H4 reacted with 2 g of H2 to produce 30 g of C2H6.
Step 3:
Determination of the limiting reactant. This can be obtained as follow:
From the balanced equation above,
28 g of C2H4 reacted with 2 g of H2.
Therefore, 9.6 g of C2H4 will react with = (9.6 x 2)/28 = 0.69 g of H2.
From the calculations made above, we can see that only 0.69 g out of 3.83 g of H2 given is needed to react completely with 9.6 g of C2H4.
Therefore, C2H4 is the limiting reactant and H2 is the excess reactant.
A. Determination of the maximum mass of ethane, C2H6 produced from the react.
In this case, the limiting reactant will be used because it will produce the maximum yield of the reaction since all of it is consumed in the reaction.
The limiting reactant is C2H4 and the maximum mass of C2H6 can be obtained as follow:
From the balanced equation above,
28 g of C2H4 reacted to produce 30 g of C2H6.
Therefore, 9.6 gof C2H4 will react to produce = (9.6 x 30)/28 = 10.29 g of C2H6.
Therefore, 10.29 g of ethane, C2H6 were produced from the reaction.
B. The limiting reactant is ethylene with formula C2H4. Please refer to step 3 above for details.
C . Determination of the mass of the excess reactant that remained after the reaction.
The excess reactant is H2, please refer to step 3 above for details and the mass that remained after the reaction can be obtained as follow:
Mass of H2 given = 3.83 g
Mass of H2 that reacted = 0.69 g
Mass of H2 remaining =.?
Mass of H2 remaining = mass of H2 given – mass of H2 that reacted.
Mass of H2 remaining = 3.83 – 0.69
Mass of H2 remaining = 3.14 g
Therefore, 3.14 g of H2 remained after the reaction.
