Hydrogen gas reactions with nitrogen to form ammonia according to the following chemical equation:
H2(g) + N2(g) --> NH3(g)
How many grams of H2 are needed to produce 12.15 grams of NH3? (express your answer as numeric only with 3 significant digits)
To determine the number of grams of H2 needed to produce 12.15 grams of NH3, we need to use stoichiometry.
The balanced equation gives us a ratio between H2 and NH3: 1 molecule of H2 reacts with 1 molecule of N2 to produce 2 molecules of NH3.
The molar mass of NH3 is 17.03 g/mol.
Now, let's calculate the molar mass of H2:
- H has a molar mass of 1.008 g/mol.
- Multiply this by the number of hydrogen atoms in one molecule of H2, which is 2: 1.008 g/mol * 2 = 2.016 g/mol.
Since the mole ratio of H2 to NH3 is 1:2, the molar mass ratio is also 1:2.
To find the number of moles of NH3 produced from 12.15 grams, we divide the mass by the molar mass:
mol NH3 = mass NH3 / molar mass NH3
mol NH3 = 12.15 g / 17.03 g/mol ≈ 0.713 mol
Since the molar mass ratio of H2 to NH3 is 1:2, the number of moles of H2 required to produce 0.713 moles of NH3 is also 0.713 moles.
Now, let's calculate the mass of H2 required:
mass H2 = moles H2 * molar mass H2
mass H2 = 0.713 mol * 2.016 g/mol ≈ 1.440 g
Therefore, approximately 1.440 grams of H2 are needed to produce 12.15 grams of NH3.
To find the number of grams of H2 needed to produce 12.15 grams of NH3, we need to use the molar ratio from the balanced chemical equation.
First, let's determine the molar masses of the compounds involved:
- H2: 2(1.00784 g/mol) = 2.01568 g/mol (rounded to 2.016 g/mol)
- NH3: 1(14.00674 g/mol) + 3(1.00784 g/mol) = 17.03052 g/mol (rounded to 17.031 g/mol)
Now, we can set up a conversion using the molar ratio from the balanced equation. The balanced equation tells us that 1 mole of H2 reacts with 1 mole of NH3. Therefore, the molar ratio of H2 to NH3 is 1:1.
Using the given mass of NH3 (12.15 g) and the molar mass of NH3 (17.031 g/mol), we can calculate the number of moles of NH3:
moles of NH3 = mass of NH3 / molar mass of NH3
moles of NH3 = 12.15 g / 17.031 g/mol ≈ 0.713 moles
Since the molar ratio is 1:1, the number of moles of H2 needed will also be 0.713 moles.
Finally, we can find the mass of H2:
mass of H2 = moles of H2 × molar mass of H2
mass of H2 = 0.713 moles × 2.016 g/mol ≈ 1.44 g
Therefore, approximately 1.44 grams of H2 are needed to produce 12.15 grams of NH3.
First, balance the equation.
3H2(g) + N2(g) → 2NH3(g)
how many moles of NH3 in 12.15 g?
You will need 3/2 that many moles of H2. Convert that to grams.