Acrylonitrile (C 3 H 3 N) is the starting material for many synthetic carpets and fabrics. It is produced by the following reaction: 2 C 3 H 6 (g) + 2 NH 3 (g) + 3 O 2 (g) → 2 C 3 H 3 N (g) + 6 H 2 O (g) If 12.0 g C 3 H 6 , 10.0 g NH 3 , and 5.0 g O 2 react, what mass of acrylonitrile can be produced, assuming 100% yield?

Acrometonitrile, which is c 3h3n, is the starting material for many synthetic carpets and fabrics. It is produced by the following reaction. Here we're told that we have 2 moles of c 3h6 reacting with 2 moles of ammonia, which is NH 3, plus 3 moles of oxygen gas to produce 2 moles of acrylonitrile plus water. Here it says, if 12 grams of c 3h6, 10 grams of ammonia, and 5 grams of oxygen gas react, what mass of acrylonitrile can be produced assuming a 100% yield? Now when they say a 100% yield at the end, that's just telling us to calculate the theoretical yield. Because remember, your theoretical yield is also your 100% yield. Here, they're giving us 3 grams of given. We're gonna have to do stoichiometry with each one of them. So we're gonna have to do stoichiometry 3 times. Alright. So we have 12 grams of c3h6 initially. Remember the stoichiometric chart. We're gonna go from grams of given to moles of given. So we have 1 mole of c3h6. On the bottom here we put the mass of c 3h6. Looking at your periodic table, 3 carbons plus 6 hydrogens. When you add them all up together, the total mass of this is 42.078 grams. So your grams of given are gone and now your moles of given. Looking at your stoichiometric chart, remember, we're now gonna go from moles of given to moles of unknown, which in this case is your acrylonitrile. When we make this transition, make this jump, we use the coefficients in the balanced equation. So it's a 2 to 2 ratio. Now, moles cancel out and now we're at moles of unknown. Finally, we change those moles of unknown into grams of unknown. And then if we add up the 3 carbons, the 3 hydrogens, and the 1 nitrogen, the combined mass is 53.069 grams. When we multiply everything on top, divide everything on the bottom, we get 15.1339 grams of acrylonitrile. One down, 2 more to go. Now we're gonna take the 10 grams of ammonia. We're gonna change those grams of given into moles of given. So one mole of ammonia, if we add up the 1 nitrogen and the 3 hydrogens, that's 17.034 grams as the total mass for n h three. Then we do the jump where we go from moles of our given ammonia to moles of our unknown, which is acrylonitrile. Look at the coefficients in the balance equation. It's also a 2 to 2 ratio. Moles cancel out. Then finally, we're gonna change the moles of acrylonitrile into grams of acrylonitrile. So this part we already did earlier, so it's just a copy and paste. Cancel out and here this would give me 31.1 539 grams of acrylonitrile. Finally, we do it for the last reactant, which is our grams of oxygen gas. So o 2 is 2 oxygens, so 1 mole of o 2 has a combined mass of 32 grams. Then we're gonna change moles of given, which is o 2 to moles of unknown, which is our acrylonitrile. Look at the balanced equation. Up here the balanced equation for every 3 moles of o 2, we have 2 moles of acrylonitrile. Then finally, we're gonna say 1 mole of acrylonitrile has a mass of 53.069 grams. So units cancel out and we'll have at the end our grams of acrylonitrile, which comes out to roughly 5.50 grams of acrylonitrile. Look at all the answers that we've obtained so far. Remember, what we do is we choose the smallest value. That smallest value represents our 100% yield or our theoretical yield. And if this is our theoretical yield, that means oxygen, the reactant that made it, has to be our limiting reactant, or limiting reagent. And that would mean that the other 2, that is n h three and c three h six, those would just represent our excess reagents. So again, this is the steps we have to take when they're giving us more than one given amount. Do stoichiometry as many times as you need to, compare your answers. The smallest answer represent your theoretical yield or your 100% yield.

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8. Quantities in Chemical Reactions

Limiting Reagent

Introduction to Chemistry

8. Quantities in Chemical Reactions

Limiting Reagent

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Multiple Choice

Acrylonitrile (C₃H₃N) is the starting material for many synthetic carpets and fabrics. It is produced by the following reaction:
2 C₃H₆ (g) + 2 NH₃ (g) + 3 O₂ (g) → 2 C₃H₃N (g) + 6 H₂O (g)
If 12.0 g C₃H₆, 10.0 g NH₃, and 5.0 g O₂ react, what mass of acrylonitrile can be produced, assuming 100% yield?

31.2 g

15.1 g

5.50 g

12.4 g

15.6 g

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Verified step by step guidance

First, determine the molar mass of each reactant: C3H6, NH3, and O2. Use the atomic masses from the periodic table: Carbon (C) = 12.01 g/mol, Hydrogen (H) = 1.01 g/mol, Nitrogen (N) = 14.01 g/mol, and Oxygen (O) = 16.00 g/mol.

Calculate the moles of each reactant using the formula: moles = mass / molar mass. For example, for C3H6, use its molar mass to find the moles from the given 12.0 g.

Identify the limiting reactant by comparing the mole ratio of the reactants to the coefficients in the balanced chemical equation. The limiting reactant is the one that will be completely consumed first, limiting the amount of product formed.

Using the moles of the limiting reactant, apply the stoichiometry of the balanced equation to calculate the moles of acrylonitrile (C3H3N) that can be produced.

Convert the moles of acrylonitrile to grams using its molar mass, which is calculated similarly to the reactants. This will give you the theoretical mass of acrylonitrile produced, assuming 100% yield.