If 0.120 g C3H6 and 0.10 g NO are allowed to mix according to the following reaction, how many g N2 are produced? 4C3H6 + 6 NO → 4C3H3N + 6H2O + N2

To determine how many grams of N2 are produced from the reaction of 0.120 g of C3H6 and 0.10 g of NO, you need to follow these steps:

Step 1: Calculate the molar mass of C3H6 and NO. The molar mass of C3H6 (propene) is calculated as follows: (3 × Carbon) + (6 × Hydrogen) = (3 × 12.01 g/mol) + (6 × 1.008 g/mol) = 36.03 g/mol + 6.048 g/mol = 42.078 g/mol.

The molar mass of NO is: (1 × Nitrogen) + (1 × Oxygen) = (1 × 14.01 g/mol) + (1 × 16.00 g/mol) = 14.01 g/mol + 16.00 g/mol = 30.01 g/mol.

Step 2: Calculate the moles of C3H6 and NO using their respective masses. For C3H6: 0.120 g C3H6 × (1 mol C3H6 / 42.078 g C3H6) ≈ 0.00285 mol C3H6 For NO: 0.10 g NO × (1 mol NO / 30.01 g NO) ≈ 0.00333 mol NO

Step 3: Determine the limiting reactant. According to the balanced equation, 4 moles of C3H6 react with 6 moles of NO. Therefore, the mole ratio of C3H6 to NO is 4:6, which simplifies to 2:3.

Now we can compare the mole ratio of the actual amounts: C3H6 has 0.00285 mol NO has 0.00333 mol

To find out if all reactants reacted completely, calculate the amount of NO that would have been needed if all of the C3H6 were to react: 0.00285 mol C3H6 × (3 mol NO / 2 mol C3H6) ≈ 0.00428 mol NO required

Since we only have 0.00333 mol of NO and we would need 0.00428 mol to completely react with the C3H6, NO is the limiting reactant.

Step 4: Calculate the moles of N2 produced using the limiting reactant (NO). According to the balanced reaction equation, 6 moles of NO produce 1 mole of N2. Since NO is the limiting reactant: 0.00333 mol NO × (1 mol N2 / 6 mol NO) ≈ 0.000555 mol N2

Step 5: Calculate the mass of N2 produced. The molar mass of N2 is 28.02 g/mol. 0.000555 mol N2 × (28.02 g N2 / 1 mol N2) ≈ 0.01555 g N2

Therefore, 0.01555 grams of N2 are produced from the reaction of 0.120 g of C3H6 and 0.10 g of NO.