Acrylonitrile (c3H3N ) is the starting material for many synthetic carpets and fabrics. It is produced by the?
following reaction:
2C3H6 (g) + 2NH3 (g) + 3O2 (G0 - 2C3H3N (g) + 6H2O (g)
If 15.0 g C3H6, 10.0 g 02, and 5.00 g Nh3 are reacted, what mass of the acrylonitrile can be produced assuming 100% yield?
----------I found the moles of c3h3n which is 56 .i do not know what to do next?
If you have mols acrylonitrile you have what you want don't you? grams = mols x molar mass. However, you don't show any work and I can't know that you did it right. Here is the way to solve the problem.
mols C3H6 = grams/molar mass.
mols NH3 = grams/molar mass.
mols O2 = grams/molar mass.
I have no idea what(G0 - 2C3H3N) means. I hope that somehow stands for an arrow but you could have written ==> a lot easier.
Now use the coefficients in the balanced equation to convert mols C3H6 to mols of the product.
Do the same for mols NH3 to mols product.
Do the same for mols O2 to mols product.
It is quite likely that the three numbers will not be the be same because this is a limiting reagent problem which means two of the values are not right. In limiting reagent problems the correct value is ALWAYS the smallest one and the reagent providing that value is the limiting reagent. Then grams of product = mols x molar mass product.
To determine the mass of acrylonitrile produced, you first need to calculate the limiting reactant. The limiting reactant is the reactant that is completely consumed and limits the amount of product that can be formed.
1. Calculate the moles of each reactant:
- moles of C3H6 = (15.0 g C3H6) / (42.08 g/mol C3H6)
- moles of NH3 = (5.00 g NH3) / (17.03 g/mol NH3)
- moles of O2 = (10.0 g O2) / (32.00 g/mol O2)
2. Divide the moles of each reactant by the coefficient in the balanced equation to find the ratio of moles:
- ratio of C3H6 = moles of C3H6 / coefficient of C3H6
- ratio of NH3 = moles of NH3 / coefficient of NH3
- ratio of O2 = moles of O2 / coefficient of O2
3. Identify the limiting reactant as the one with the smallest ratio. This reactant will be completely consumed.
4. Use the limiting reactant to calculate the moles of acrylonitrile produced:
- moles of C3H3N = ratio of limiting reactant / coefficient of C3H3N
5. Multiply the moles of C3H3N by the molar mass of acrylonitrile to get the mass of acrylonitrile produced.
Remember to round your final answer to the appropriate number of significant figures.
To find the mass of acrylonitrile that can be produced, you need to use the given mole ratios from the balanced equation and convert the moles of C3H3N to grams.
From the balanced equation, we can see that the mole ratio between C3H3N and C3H6 is 1:2. This means that for every 2 moles of C3H6, we will get 1 mole of C3H3N.
Given that you found the moles of C3H3N to be 56, this means that there must have been 2x moles of C3H6, where x is the number of moles of C3H3N.
So, if we divide 56 by 2, we get x = 28, which is the number of moles of C3H6.
Next, you need to determine the limiting reactant. The limiting reactant is the reactant that is completely consumed in the reaction, limiting the amount of product that can be formed.
To find the limiting reactant, you can compare the moles of each reactant to their respective coefficient in the balanced equation.
For C3H6:
Moles of C3H6 = mass/molar mass = 15.0 g / 42.08 g/mol = 0.357 mol
For NH3:
Moles of NH3 = mass/molar mass = 5.00 g / 17.03 g/mol = 0.293 mol
For O2:
Moles of O2 = mass/molar mass = 10.0 g / 32.00 g/mol = 0.313 mol
Comparing the moles of the reactants to their coefficients:
C3H6:NH3:O2 = 0.357:0.293:0.313
From the ratio, we can see that there are fewer moles of NH3 compared to C3H6, meaning NH3 is the limiting reactant.
Now, we need to find how many moles of C3H3N can be formed from 0.293 moles of NH3.
From the balanced equation, the mole ratio between NH3 and C3H3N is 2:1. So, if we have 0.293 moles of NH3, we will get half that amount in moles of C3H3N.
Moles of C3H3N = 0.293/2 = 0.147 mol
Finally, we can calculate the mass of acrylonitrile produced.
Mass of C3H3N = moles of C3H3N x molar mass of C3H3N
= 0.147 mol x 53.06 g/mol
= 7.829 g
Therefore, assuming 100% yield, 7.829 grams of acrylonitrile can be produced.