a)How many grams of H2 are needed to produce 14.27 g of NH3?
b)How many molecules (not moles) of NH3 are produced from 1.57×10−4 g of H2 ?
You need an equation, then follow the steps in this example problem.
http://www.jiskha.com/science/chemistry/stoichiometry.html
if you have 150g of nano3 how many mole are there
To find the amount of H2 required to produce 14.27 g of NH3, we need to follow the balanced chemical equation for the reaction between H2 and NH3.
a) Balanced equation: N2 + 3H2 -> 2NH3
From the equation, we can see that 3 moles of H2 are required to produce 2 moles of NH3.
Step 1: Convert grams of NH3 to moles of NH3.
Using the molar mass of NH3 (17.03 g/mol), we can calculate the moles of NH3.
moles of NH3 = mass of NH3 / molar mass of NH3
moles of NH3 = 14.27 g / 17.03 g/mol
moles of NH3 = 0.8373 mol
Step 2: Convert moles of NH3 to moles of H2.
Since the molar ratio of H2 to NH3 is 3:2, we can calculate the moles of H2.
moles of H2 = moles of NH3 * (3/2)
moles of H2 = 0.8373 mol * (3/2)
moles of H2 = 1.25595 mol
Step 3: Convert moles of H2 to grams of H2.
Using the molar mass of H2 (2.02 g/mol), we can calculate the grams of H2 required.
grams of H2 = moles of H2 * molar mass of H2
grams of H2 = 1.25595 mol * 2.02 g/mol
grams of H2 = 2.53519 g
Therefore, to produce 14.27 g of NH3, you would need approximately 2.54 grams of H2.
b)To find the number of molecules of NH3 produced from 1.57x10^-4 g of H2, we will follow the same steps but convert grams of H2 to moles of H2 instead.
Step 1: Convert grams of H2 to moles of H2.
Using the molar mass of H2 (2.02 g/mol), we can calculate the moles of H2.
moles of H2 = mass of H2 / molar mass of H2
moles of H2 = 1.57x10^-4 g / 2.02 g/mol
moles of H2 = 7.76x10^-5 mol
Step 2: Convert moles of H2 to moles of NH3.
Since the molar ratio of H2 to NH3 is 3:2, we can calculate the moles of NH3.
moles of NH3 = moles of H2 * (2/3)
moles of NH3 = 7.76x10^-5 mol * (2/3)
moles of NH3 = 5.18x10^-5 mol
Step 3: Convert moles of NH3 to molecules of NH3.
Using Avogadro's number (6.022x10^23 molecules/mol), we can calculate the molecules of NH3.
molecules of NH3 = moles of NH3 * Avogadro's number
molecules of NH3 = 5.18x10^-5 mol * 6.022x10^23 molecules/mol
molecules of NH3 = 3.12x10^19 molecules
Therefore, 1.57x10^-4 g of H2 would produce approximately 3.12x10^19 molecules of NH3.
To answer these questions, we need to use stoichiometry, which is the relationship between the amounts of reactants and products in a chemical reaction.
a) To find out how many grams of H2 are needed to produce 14.27 g of NH3, we need to know the balanced chemical equation for the reaction. The balanced equation for the formation of NH3 from H2 is:
3H2 + N2 → 2NH3
According to this equation, 3 moles of H2 are required to produce 2 moles of NH3. To find the grams of H2 needed, we can set up a proportion.
First, calculate the molar mass (grams per mole) of NH3:
NH3 = (1 x 1.01 g/mol H) + (3 x 1.01 g/mol H) = 17.03 g/mol
Next, set up the proportion:
(3 moles H2 / 2 moles NH3) = (x grams H2 / 14.27 g NH3)
Now, solve for x to find the grams of H2 needed:
x = (3 moles H2 / 2 moles NH3) * (14.27 g NH3 / 1)
= (3/2) * 14.27 g
= 21.405 g
So, 21.405 grams of H2 are needed to produce 14.27 g of NH3.
b) To find out how many molecules of NH3 are produced from 1.57×10^-4 g of H2, we first need to calculate the number of moles of H2.
First, calculate the molar mass of H2:
H2 = (2 x 1.01 g/mol H) = 2.02 g/mol
Next, use the formula:
moles = mass / molar mass
moles H2 = 1.57×10^-4 g / 2.02 g/mol ≈ 7.8×10^-5 mol
Now, we use the balanced chemical equation to find the molar ratio between H2 and NH3. From the equation 3H2 + N2 → 2NH3, we see that 3 moles of H2 produce 2 moles of NH3.
Using this ratio, we can calculate the number of moles of NH3 produced:
moles NH3 = (7.8×10^-5 mol H2) * (2 mol NH3 / 3 mol H2)
= 5.2×10^-5 mol NH3
Finally, to find the number of molecules, we multiply the number of moles by Avogadro's number (6.022×10^23 molecules/mol):
molecules NH3 = (5.2×10^-5 mol NH3) * (6.022×10^23 molecules/mol)
≈ 3.13×10^19 molecules NH3
So, approximately 3.13×10^19 molecules of NH3 are produced from 1.57×10^-4 g of H2.