how many grams of ethylene (C2H4) are required for the complete reaction of 31.7 grams of hydrogen gas?
C2H4 + H2 ==> C2H6
mols H2 gas = 31.7/molar mass.
1 mol H2 = 1 mol C2H4 from the coefficients in the balanced equation.
g C2H2 = mols C2H4 x molar mass C2H4
To determine the amount of grams of ethylene (C2H4) required for the complete reaction of 31.7 grams of hydrogen gas, we first need to establish a balanced chemical equation for the reaction.
The balanced chemical equation for the reaction between ethylene (C2H4) and hydrogen gas (H2) is as follows:
C2H4 + H2 -> C2H6
This equation shows that one molecule of ethylene reacts with one molecule of hydrogen gas to form one molecule of ethane (C2H6).
Next, we need to determine the molar mass of ethylene (C2H4) and hydrogen gas (H2).
The molar mass of ethylene (C2H4) can be calculated as follows:
C = 12.01 g/mol
H = 1.01 g/mol
Molar mass of C2H4 = (2 x C) + (4 x H) = (2 x 12.01 g/mol) + (4 x 1.01 g/mol) = 28.05 g/mol
The molar mass of hydrogen gas (H2) is simply 2.02 g/mol since there are two hydrogen atoms in each molecule.
Now we can calculate the number of moles of hydrogen gas (H2) using its given mass:
Moles of H2 = Mass of H2 / Molar mass of H2
= 31.7 g / 2.02 g/mol
≈ 15.69 mol
Since the balanced equation shows a 1:1 stoichiometric ratio between ethylene and hydrogen gas, the number of moles of ethylene required will also be 15.69 mol.
Finally, we can calculate the grams of ethylene required:
Grams of C2H4 = Moles of C2H4 × Molar mass of C2H4
= 15.69 mol × 28.05 g/mol
≈ 439.7 g
Therefore, approximately 439.7 grams of ethylene (C2H4) are required for the complete reaction of 31.7 grams of hydrogen gas.