I think I managed to totally screw up my computer so I did already post this question, but from what I see it is gone now. I don't really know if its actually there or not cuz I can't do anything with computers but here is the question:
Hydrogen cyanide is produced industrially from the reaction of gaseous ammonia, oxygen, and methane.
2 NH3(g) + 3 O2(g) + 2 CH4(g) �¨ 2 HCN(g) + 6 H2O(g)
If 5.27x10^3 kg each of NH3, O2, and CH4 are reacted, what mass of H2O will be produced, assuming 100% yield?
You have the balanced equation:
first figure the amount of moles of ammonia, oxygen, and methane in the mass used. You need 1.5 times the methane amount of Oxygen.
So which is the limiting reactant? Just looking at it, I suspect it is O2, (working it in my head).
once I have the limiting reactant how do I go about calculating the mass of h20 produced? Dimensional Analysis would be a good way for me to understand it.
Sorry, I hit the wrong button.
Dimensional analysis will work. Just use the coefficients of the balanced equation.
For example,
mols starting material x (coefficient*product desired/coefficient*starting matrial) = mols product desired.
To determine the mass of H2O produced, we need to first calculate the limiting reactant and then use stoichiometry to find the mass of water produced.
1. Calculate the number of moles for each reactant:
- NH3 (Ammonia): 5.27x10^3 kg
- O2 (Oxygen): 5.27x10^3 kg
- CH4 (Methane): 5.27x10^3 kg
To convert the mass of each reactant to moles, we need to use their molar masses.
The molar mass of NH3 is: 17 g/mol.
The molar mass of O2 is: 32 g/mol.
The molar mass of CH4 is: 16 g/mol.
Moles of NH3 = (5.27x10^3 kg) / (17 g/mol)
Moles of O2 = (5.27x10^3 kg) / (32 g/mol)
Moles of CH4 = (5.27x10^3 kg) / (16 g/mol)
2. Determine the limiting reactant:
To find the limiting reactant, we compare the ratios of the coefficients in the balanced equation to the moles of each reactant.
From the balanced equation:
2 NH3 : 3 O2 : 2 CH4
2 : 3 : 2
The limiting reactant is the one that produces the least amount of product based on the balanced equation's stoichiometry.
Let's compare the ratios:
Moles of NH3 / Coefficient of NH3 = Moles of NH3 / 2 = x
Moles of O2 / Coefficient of O2 = Moles of O2 / 3 = y
Moles of CH4 / Coefficient of CH4 = Moles of CH4 / 2 = z
Calculate x, y, and z using the values obtained earlier.
3. Determine the amount of H2O produced from the limiting reactant:
The balanced equation states that from 2 moles of NH3, 6 moles of H2O are produced.
Since the coefficient ratio of NH3 to H2O is 2:6 (or simplified to 1:3), we can use this ratio to calculate the moles of H2O that would be produced if NH3 were the limiting reactant.
Moles of H2O = x * 3
4. Convert moles of H2O to mass:
Now, multiply the moles of H2O by the molar mass of water to get the mass of water.
Mass of H2O = Moles of H2O * Molar mass of H2O
The molar mass of H2O is 18 g/mol.
Finally, compute the mass of H2O produced!
Note: Since the question assumes 100% yield, we can use the stoichiometry directly without considering any limiting reactants.