For the following reaction, 4.35 grams of hydrogen gas are allowed to react with with 10.3 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
To find the maximum amount of ethane (C2H6) that can be formed, we need to determine the limiting reagent between hydrogen gas (H2) and ethylene (C2H4).
1. Start by calculating the number of moles for each reactant.
- Moles of H2 = mass of H2 / molar mass of H2
- Moles of C2H4 = mass of C2H4 / molar mass of C2H4
2. Next, use the mole ratios from the balanced equation to determine the number of moles of ethane (C2H6) that can be formed from each reactant.
- Moles of C2H6 from H2 = moles of H2 * (2 moles of C2H6 / 2 moles of H2)
- Moles of C2H6 from C2H4 = moles of C2H4 * (1 mole of C2H6 / 1 mole of C2H4)
3. Compare the moles of ethane (C2H6) calculated from each reactant to find the limiting reagent.
- The limiting reagent is the one that produces the lower number of moles of C2H6.
4. Use the limiting reagent to calculate the maximum amount of ethane (C2H6) that can be formed.
- Multiply the moles of C2H6 from the limiting reagent by the molar mass of C2H6 to get the maximum mass of C2H6 that can be formed.
To find the formula for the limiting reagent, compare the coefficients in front of the reactants in the balanced equation. The limiting reagent will be the one with a coefficient of 1.
To find the amount of the excess reagent that remains after the reaction is complete, subtract the amount of the limiting reagent that reacted from the initial amount of the excess reagent. Calculate the moles of the excess reagent using its molar mass and convert it back to grams if needed.
To determine the maximum amount of ethane (C2H6) that can be formed, we need to identify the limiting reagent. The limiting reagent is the reactant that is completely consumed and determines the maximum amount of product that can be formed.
Step 1: Calculate the moles of each reactant.
- Hydrogen (H2): Given mass = 4.35 grams, molar mass = 2g/mol
Number of moles of H2 = mass/molar mass = 4.35g/2g/mol = 2.175 mol
- Ethylene (C2H4): Given mass = 10.3 grams, molar mass = 28g/mol
Number of moles of C2H4 = mass/molar mass = 10.3g/28g/mol = 0.367 mol
Step 2: Calculate the moles of ethane (C2H6) that can be formed from each reactant based on their stoichiometry.
From the balanced equation, the mole ratio between hydrogen (H2) and ethane (C2H6) is 1:1, and the mole ratio between ethylene (C2H4) and ethane (C2H6) is 1:1.
- Moles of ethane (C2H6) formed from H2 = 2.175 mol * (1 mol C2H6/1 mol H2) = 2.175 mol
- Moles of ethane (C2H6) formed from C2H4 = 0.367 mol * (1 mol C2H6/1 mol C2H4) = 0.367 mol
Step 3: Determine the limiting reagent.
The limiting reagent is the reactant that forms the least amount of product. In this case, the reactant that forms the least amount of ethane (C2H6) is ethylene (C2H4), with a production of 0.367 mol.
Step 4: Calculate the maximum amount of ethane (C2H6) that can be formed.
The molar mass of ethane (C2H6) is 30g/mol.
- Mass of ethane (C2H6) formed from C2H4 = 0.367 mol * 30g/mol = 11.01 grams
Therefore, the maximum amount of ethane (C2H6) that can be formed is 11.01 grams.
To determine the formula of the limiting reagent, we can compare the given amount of each reactant with their respective molar masses.
- Hydrogen (H2) molar mass = 2g/mol
Given mass of H2 = 4.35 grams > Molar mass of H2
- Ethylene (C2H4) molar mass = 28g/mol
Given mass of C2H4 = 10.3 grams > Molar mass of C2H4
Since both reactants have masses greater than their respective molar masses, neither is in excess. Therefore, the formula of the limiting reagent cannot be determined.
Finally, to determine the amount of excess reagent remaining after the reaction is complete, we need to calculate the moles of the excess reagent left.
- Moles of excess H2 remaining = Moles of H2 initially - Moles of H2 used in the reaction
= 2.175 mol - 0.367 mol (used in the reaction) = 1.808 mol
To convert the moles of excess H2 remaining to grams, multiply it by the molar mass of H2:
- Mass of excess H2 remaining = 1.808 mol * 2g/mol = 3.616 grams
Therefore, the amount of excess H2 that remains after the reaction is complete is 3.616 grams.