In the following reaction, how many grams of ammonia (NH3) will produce 300 grams of N2?
4NH3 + 6NO -> 5N2 + 6H2O
Each 4 moles of NH3 produce 5 moles of N2
so, you get
4/5 * 300/14 * 17.03 = 292g NH3
To determine the grams of ammonia (NH3) required to produce 300 grams of N2, you need to use the stoichiometry of the reaction. The coefficients in the balanced equation represent the mole ratios between the different substances involved in the reaction.
In this reaction, the coefficient in front of NH3 is 4, indicating that for every 4 moles of NH3, you get 5 moles of N2. Therefore, the mole ratio of NH3 to N2 is 4:5.
Here's the step-by-step process to calculate the grams of NH3:
Step 1: Find the molar mass of NH3.
The molecular weight of nitrogen (N) is 14.01 grams per mole, and the molecular weight of hydrogen (H) is 1.01 grams per mole. Since there are three hydrogen atoms in ammonia, the molar mass of NH3 is:
(14.01 g/mol for N) + (3 * 1.01 g/mol for H) = 17.03 g/mol
Step 2: Calculate the moles of N2.
Divide the given mass of N2 (300 g) by its molar mass to find the moles:
moles of N2 = (mass of N2) / (molar mass of N2)
moles of N2 = 300 g / (28.02 g/mol for N2) = 10.707 mol
Step 3: Determine the moles of NH3 using the mole ratio.
Since the mole ratio of NH3 to N2 is 4:5, multiply the moles of N2 by the ratio to find the moles of NH3:
moles of NH3 = (moles of N2) * (4/5)
moles of NH3 = 10.707 mol * (4/5) = 8.5664 mol
Step 4: Convert moles of NH3 to grams.
Multiply the moles by the molar mass of NH3 to convert to grams:
grams of NH3 = (moles of NH3) * (molar mass of NH3)
grams of NH3 = 8.5664 mol * (17.03 g/mol for NH3) = 145.96 g
Therefore, to produce 300 grams of N2, you would need approximately 145.96 grams of NH3.