If 15.0 grams of C3H6, 10.0 grams of oxygen, and 5.00 grams of NH3 are reacted, what mass of acrylonitrile
can be produced assuming 100% yield, what mass of the excess reactants remain?
15g C3H6 (1 mol / 42.08 g) = .35 mol C3H6
.36 mol C3H6 ( 2 mol C3H3N / 2 mol C3H6) = .36 mol C3H3N
10g O2 ( 1 mol/ 32 g O2) = .313 mol O2
.313 mol O2 (2 mol C3H3N/ 3 mol O2) -= .209 mol C3H3N which is the limiting reactant
5 g NH3 ( 1 mol NH3/ 17.03 g) = .294 mol C3H3N
.209 mol C3H3N ( 32 c3H6 / 2 mol C3H3N) = .209 m C3H6
.36mol - .209 mol = .151 mol C3H6
.151 mol C3H6 (42.08 g / 1 mol C3H6 ) = 6.35 g C3H6
Thank you for your help!!
You're welcome! I'm here to provide some extra amusement while helping you out. So, with a 100% yield, you can produce approximately 6.35 grams of acrylonitrile. That's quite a handful! Now, for the excess reactants, after all the reaction shenanigans, you're left with about 0.151 moles of C3H6. That's like the stragglers who couldn't keep up with the acrylonitrile party. Converted to weight, it weighs around 5.00 grams. So, these excess reactants just couldn't resist sticking around for some more fun!
Based on the given reactants and assuming 100% yield, the mass of acrylonitrile that can be produced is 6.35 grams. The mass of excess reactants remaining is 0.151 mol of C3H6, which is equivalent to 6.35 grams.
To find the mass of acrylonitrile that can be produced, we need to determine the limiting reactant. The limiting reactant is the reactant that is completely consumed in the reaction and therefore determines the amount of product that can be formed.
Given:
Mass of C3H6 = 15.0g
Mass of O2 = 10.0g
Mass of NH3 = 5.00g
Step 1: Convert the masses to moles.
The molar mass of C3H6 is 42.08 g/mol.
The molar mass of O2 is 32 g/mol.
The molar mass of NH3 is 17.03 g/mol.
15.0g C3H6 × (1 mol / 42.08 g) = 0.356 mol C3H6
10.0g O2 × (1 mol / 32 g) = 0.313 mol O2
5.00g NH3 × (1 mol / 17.03 g) = 0.294 mol NH3
Step 2: Calculate the moles of acrylonitrile that can be produced from each reactant.
For C3H6:
0.356 mol C3H6 × (2 mol C3H3N / 2 mol C3H6) = 0.356 mol C3H3N
For O2:
0.313 mol O2 × (2 mol C3H3N / 3 mol O2) = 0.209 mol C3H3N
The moles of acrylonitrile formed from NH3 is not needed since it is not the limiting reactant.
Step 3: Determine the limiting reactant.
The reactant that produces a smaller amount of product is the limiting reactant. In this case, O2 produces 0.209 mol C3H3N, which is smaller than the 0.356 mol C3H3N produced by C3H6. Therefore, O2 is the limiting reactant.
Step 4: Calculate the mass of acrylonitrile that can be produced.
The molar mass of C3H3N is 53.06 g/mol.
0.209 mol C3H3N × (53.06 g / 1 mol C3H3N) = 11.08 g C3H3N
Therefore, the mass of acrylonitrile that can be produced is 11.08 grams.
To find the mass of the excess reactants remaining, we need to subtract the moles of the limiting reactant used from the initial moles of each reactant.
For C3H6:
0.356 mol C3H6 - 0.209 mol C3H6 = 0.147 mol C3H6 remaining
0.147 mol C3H6 × (42.08 g / 1 mol C3H6) = 6.19 g C3H6 remaining
For NH3:
0.294 mol NH3 - 0 mol NH3 (since NH3 is not the limiting reactant) = 0.294 mol NH3 remaining
0.294 mol NH3 × (17.03 g / 1 mol NH3) = 5.00 g NH3 remaining
Therefore, the mass of the excess reactants remaining is 6.19 grams of C3H6 and 5.00 grams of NH3.