When you react hydrogen gas with nitrogen gas to form ammonia, How many grams of all species ( reactants and products) are left when you start with 140 grams of nitrogen and 40 grams of hydrogen?
Follow the instructions for the above two questions.
3.5g
To answer this question, we need to determine the limiting reactant between hydrogen gas (H2) and nitrogen gas (N2). The limiting reactant is the reactant that is completely consumed and determines the amount of product that can be formed.
To find the limiting reactant, we can use the balanced chemical equation for the reaction:
N2 + 3H2 → 2NH3
First, we need to calculate the number of moles for each reactant. The molar mass of nitrogen (N2) is 28.02 g/mol, and the molar mass of hydrogen (H2) is 2.02 g/mol.
Number of moles of nitrogen = mass of nitrogen / molar mass of nitrogen
Number of moles of nitrogen = 140 g / 28.02 g/mol ≈ 4.993 mol
Number of moles of hydrogen = mass of hydrogen / molar mass of hydrogen
Number of moles of hydrogen = 40 g / 2.02 g/mol ≈ 19.802 mol
According to the balanced equation, the reaction requires 3 moles of hydrogen for every 1 mole of nitrogen. Therefore, we need to compare the stoichiometric ratios of the reactants:
Ratio of nitrogen to hydrogen = 4.993 mol N2 / 3 mol H2 ≈ 1.664
Ratio of nitrogen to hydrogen = 19.802 mol H2 / 1 mol H2 ≈ 19.802
Since the ratio of nitrogen to hydrogen is lower when comparing the reactants, hydrogen is the limiting reactant. This means that all the hydrogen will be consumed before nitrogen is completely used up.
Now, we can calculate the amount of ammonia (NH3) produced using the stoichiometry of the balanced equation. The ratio of ammonia to hydrogen is 2:3.
Number of moles of ammonia = (Number of moles of hydrogen) x (2 mol NH3 / 3 mol H2)
Number of moles of ammonia = (19.802 mol H2) x (2 mol NH3 / 3 mol H2) ≈ 13.201 mol
Finally, we can calculate the mass of each species (reactants and products) using their respective molar masses.
Mass of nitrogen remaining = (Number of moles of nitrogen) x (molar mass of nitrogen)
Mass of nitrogen remaining = 4.993 mol x 28.02 g/mol ≈ 139.942 g
Mass of hydrogen remaining = 0 g (since it is completely consumed)
Mass of ammonia formed = (Number of moles of ammonia) x (molar mass of ammonia)
Mass of ammonia formed = 13.201 mol x 17.03 g/mol ≈ 224.843 g
Therefore, when starting with 140 grams of nitrogen and 40 grams of hydrogen, there will be approximately 139.942 grams of nitrogen remaining, 0 grams of hydrogen remaining, and 224.843 grams of ammonia formed.