Acrylonitrile (c3h3n) is the starting material for many synthetic carpets and fabrics.

15.0 g c3h6, 10.0 g o2, and 5.00 g nh3 are reacted to produce acrylonitrile and water.

_c3h6 + __o2 + __nh3 → + __h2o

what is the balanced equation?

what is the limiting reactant?

how much of the excessive reactant is left over?

how much of each product is produced?

Q1: The balanced equation: C₃H₆ + 3/2O₂ + NH₃ → C₃H₃N + 3H₂O.b

Q2: The limiting reactants is O₂.

Q3: mass of C₃H₆ excessive = 6.21 g.

mass of NH₃ excessive = 1.462 g.

Q4: The mass of C₃H₃N = 11.036 g.

the mass of H₂O = no. of moles x molar mass = (0.208 mol)(18.0 g/mol) = 3.744 g.

Explanation:

Q1: What is the balanced equation?

To balance the equation, you should apply the law of conversation of mass that the no. of atoms in the reactants side is equal to the no. of atoms in the products side.

So, cc

It is clear that 1.0 mole of C₃H₆ reacts with 1.5 mole of O₂ and 1.0 mole of NH₃ to produce 1.0 mole of C₃H₃N and 3.0 moles H₂O.

Q2: What is the limiting reactant?

To determine the limiting reactant, we should calculate the no. of moles of each reactant firstly:

no. of moles of C₃H₆ = mass/molar mass = (15.0 g)/(42.08 g/mol) = 0.356 mol.

no. of moles of O₂ = mass/molar mass = (10.0 g)/(32.0 g/mol) = 0.3125 mol.

no. of moles of NH₃ = mass/molar mass = (5.0 g)/(17.0 g/mol) = 0.294 mol.

The limiting reactant is the reactant that has the lowest mole ratio to other reactants.

From the balanced equation: The ratio of the reactants to each other C₃H₆: O₂: NH₃ is (1: 1.5: 1).

So, 0.3125 reacts completely with 0.208 mole of both C₃H₆ and NH₃ with (1.5: 1: 1) ratio, and the remaining of them is in excess.

Thus, the limiting reactants is O₂.

Q3: How much of the excessive reactant is left over?

The excessive reactants are C₃H₆ and NH₃.

The remaining no. of moles of C₃H₆ = 0.356 mol - 0.208 mol = 0.1476 mol.

The remaining no. of moles of NH₃ = 0.294 mol - 0.208 mol = 0.086 mol.

Now, we can get the excessive mass:

mass of C₃H₆ excessive = no. of moles  x molar mass = (0.1476 mol)(42.08 g/mol) = 6.21 g.

mass of NH₃ excessive = no. of moles  x molar mass = (0.086 mol)(17.0 g/mol) = 1.462 g.

Q4: How much of each product is produced?

We should get the no. of moles of the produced C₃H₃N + 3H₂O.

Using cross multiplication:

1.0 mole of C₃H₆ produces → 1.0 mole of C₃H₃N.

0.208 mole of C₃H₆ produces → ??? mole of C₃H₃N.

∴ The no. of moles of C₃H₃N is 0.208.

So, the mass of C₃H₃N = no. of moles x molar mass = (0.208 mol)(53.06 g/mol) = 11.036 g.

Also, for water:

Using cross multiplication:

1.0 mole of C₃H₆ produces → 1.0 mole of H₂O.

0.208 mole of C₃H₆ produces → ??? mole of H₂O.

∴ The no. of moles of H₂O is 0.208.

So, the mass of H₂O = no. of moles x molar mass = (0.208 mol)(18.0 g/mol) = 3.744 g.