Final answer:
To find the number of grams of N2 formed from the reaction of 21 grams of NH3, use stoichiometry to convert the mass of NH3 to moles, use the mole ratio from the balanced equation to find the moles of N2 produced, and then convert those moles back to grams, resulting in approximately 17.23 grams of N2.
Step-by-step explanation:
To calculate the number of grams of N2 formed from the reaction of 21 grams of NH3, we can use stoichiometry. First, we calculate the molar mass of NH3, which is about 17.03 g/mol. We then determine how many moles of NH3 the 21 grams represent:
(21 g NH3) ÷ (17.03 g/mol NH3) = 1.23 mol NH3
Looking at the balanced equation, 4NH3 + 3O2 → 2N2 + 6H2O, it shows that 4 moles of NH3 yield 2 moles of N2. We can set up the ratio:
(2 mol N2) / (4 mol NH3) = (x mol N2) / (1.23 mol NH3)
Solving for x gives us 0.615 mol N2 (since 1.23 is approximately a quarter of 4, half of that would be approximately a quarter of 2). The molar mass of N2 is approximately 28.02 g/mol, so we multiply the moles of N2 by the molar mass to find the grams of N2 formed:
(0.615 mol N2) × (28.02 g/mol) ≈ 17.23 g N2
Therefore, the reaction of 21 grams of NH3 would result in the formation of approximately 17.23 grams of N2.