this is confusing me so bad!!
Nitrogen gas reacts with hydrogen gas to produce ammonia by the following equation:
N2(g)+3H2(g)→2NH3(g)
If you have 5.79g of H2, how many grams of NH3 can be produced?
I said, 2/3 *5.79 = 3.86 but that's not the answer.. it says the answer is 32.5 g? how??
How many grams of H2 are needed to react with 2.30 g N2? Do I need to find moles or something?? I was trying to mole ratio then times the mass of element?
nevermind, I figured it out... ughhhhh
You're still trying to make grams go to grams. You CAN'T do that. You MUST change grams to mols, convert mols of what you have to mols of what you want, then change back to grams.
To solve these types of problems, we need to use stoichiometry, which involves converting the given mass of one substance to the mass of another substance involved in a chemical reaction.
Let's start with the first question:
If you have 5.79 g of H2, and you want to determine how many grams of NH3 can be produced, you need to use the stoichiometric ratio provided by the balanced chemical equation.
The balanced equation tells us that for every 3 moles of H2, we produce 2 moles of NH3. So we need to convert the mass of H2 to moles, then use the stoichiometric ratio to convert to moles of NH3, and finally convert back to grams of NH3.
1. Calculate the number of moles of H2:
To do this, divide the given mass of H2 by its molar mass.
Molar mass of H2 = 2.016 g/mol (2 hydrogen atoms per mole)
moles of H2 = 5.79 g H2 / 2.016 g/mol ≈ 2.87 mol H2
2. Apply the stoichiometric ratio using the molar ratio from the balanced equation:
From the balanced equation, we know that for every 3 moles of H2, we produce 2 moles of NH3.
moles of NH3 = (2.87 mol H2) x (2 mol NH3 / 3 mol H2) ≈ 1.91 mol NH3
3. Convert moles of NH3 to grams of NH3:
To do this, multiply the number of moles of NH3 by its molar mass.
Molar mass of NH3 = 17.031 g/mol (1 nitrogen atom + 3 hydrogen atoms)
mass of NH3 = 1.91 mol NH3 x 17.031 g/mol ≈ 32.51 g NH3
So, the correct answer is approximately 32.5 g of NH3, not 3.86 g as you initially calculated.
Now, let's move on to the second question:
You are correct in using the mole ratio and mass of the element involved.
Given: 2.30 g of N2
We want to find the mass of H2 needed to react with this given mass of N2.
1. Convert the mass of N2 to moles, using the molar mass of N2:
Molar mass of N2 = 28.014 g/mol (2 nitrogen atoms)
moles of N2 = 2.30 g N2 / 28.014 g/mol ≈ 0.082 mol N2
2. Apply the stoichiometric ratio:
From the balanced equation, we know that for every 1 mole of N2, we need 3 moles of H2.
moles of H2 = (0.082 mol N2) x (3 mol H2 / 1 mol N2) ≈ 0.247 mol H2
3. Convert moles of H2 to grams of H2:
Molar mass of H2 = 2.016 g/mol (2 hydrogen atoms)
mass of H2 = 0.247 mol H2 x 2.016 g/mol ≈ 0.498 g H2
Therefore, you would need approximately 0.498 g of H2 to react with 2.30 g of N2.
Remember, stoichiometry allows us to relate the amounts of substances in a chemical reaction based on a balanced equation, by using moles as the conversion factor.