If 5.00 103 kg each of NH3, O2, and CH4 are reacted, what mass of HCN and of H2O will be produced, assuming 100% yield?
2 NH3(g) + 3 O2(g) + 2 CH4(g) 2 HCN(g) + 6 H2O(g)
You don't have an equation. Where is the arrow? I can't tell the reactants from the products.
the equation goes as follows:
2 NH3(g) + 3 O2(g) + 2 CH4(g) ---> 2 HCN(g) + 6 H2O(g)
I need help with this, also!
To find the mass of HCN and H2O produced, we need to first calculate the number of moles of NH3, O2, and CH4 that are reacted.
Given:
Mass of NH3 = 5.00 x 10^3 kg
Mass of O2 = 5.00 x 10^3 kg
Mass of CH4 = 5.00 x 10^3 kg
We can calculate the number of moles using the molar mass of each compound:
Molar mass of NH3: 1 x 14.01 g/mol + 3 x 1.01 g/mol = 17.03 g/mol
Molar mass of O2: 2 x 16.00 g/mol = 32.00 g/mol
Molar mass of CH4: 1 x 12.01 g/mol + 4 x 1.01 g/mol = 16.05 g/mol
Number of moles of NH3 = Mass of NH3 / Molar mass of NH3
= (5.00 x 10^3 kg) / (17.03 g/mol)
= (5.00 x 10^6 g) / (17.03 g/mol)
= 2.936 x 10^5 mol
Number of moles of O2 = Mass of O2 / Molar mass of O2
= (5.00 x 10^3 kg) / (32.00 g/mol)
= (5.00 x 10^6 g) / (32.00 g/mol)
= 1.563 x 10^5 mol
Number of moles of CH4 = Mass of CH4 / Molar mass of CH4
= (5.00 x 10^3 kg) / (16.05 g/mol)
= (5.00 x 10^6 g) / (16.05 g/mol)
= 3.113 x 10^5 mol
According to the balanced chemical equation, the reaction requires 2 moles of NH3, 3 moles of O2, and 2 moles of CH4 to produce 2 moles of HCN and 6 moles of H2O.
By stoichiometry, we can determine the moles of HCN and H2O produced.
Number of moles of HCN = (Number of moles of NH3 / 2) = (2.936 x 10^5 mol / 2) = 1.468 x 10^5 mol
Number of moles of H2O = (Number of moles of NH3 x 6) = (2.936 x 10^5 mol x 6) = 1.762 x 10^6 mol
Finally, we can calculate the mass of HCN and H2O produced using their molar masses.
Mass of HCN = Number of moles of HCN x Molar mass of HCN
= (1.468 x 10^5 mol) x [(1 x 1.01 g/mol) + (1 x 12.01 g/mol) + (1 x 14.01 g/mol)]
= 4.204 x 10^5 g
= 420.4 kg
Mass of H2O = Number of moles of H2O x Molar mass of H2O
= (1.762 x 10^6 mol) x [(2 x 1.01 g/mol) + (1 x 16.00 g/mol)]
= 5.652 x 10^6 g
= 5652 kg
Therefore, assuming 100% yield, the mass of HCN produced is 420.4 kg, and the mass of H2O produced is 5652 kg.
To find the mass of HCN and H2O produced, we need to use stoichiometry and the given molar ratios of the reactants and products.
First, let's calculate the number of moles for each reactant based on their given mass:
Mass of NH3 = 5.00 x 10^3 kg
Molar mass of NH3 = 17.03 g/mol
Number of moles of NH3 = (5.00 x 10^3 kg) / (17.03 g/mol)
Similarly, calculate the number of moles for O2 and CH4:
Mass of O2 = 5.00 x 10^3 kg
Molar mass of O2 = 32.00 g/mol
Number of moles of O2 = (5.00 x 10^3 kg) / (32.00 g/mol)
Mass of CH4 = 5.00 x 10^3 kg
Molar mass of CH4 = 16.04 g/mol
Number of moles of CH4 = (5.00 x 10^3 kg) / (16.04 g/mol)
Now, let's determine the limiting reactant. The limiting reactant is the reactant that is completely consumed and determines the maximum amount of product that can be formed.
The balanced equation shows that to produce 2 moles of HCN, we need 2 moles of NH3, 3 moles of O2, and 2 moles of CH4. Therefore, we need to compare the moles of each reactant to find the limiting reactant.
Comparing the moles of each reactant:
Moles of NH3 = (5.00 x 10^3 kg) / (17.03 g/mol)
Moles of O2 = (5.00 x 10^3 kg) / (32.00 g/mol)
Moles of CH4 = (5.00 x 10^3 kg) / (16.04 g/mol)
Next, divide the moles of each reactant by the stoichiometric coefficient to determine how many moles of product can be formed for each reactant:
Moles of HCN produced from NH3 = (Moles of NH3) / 2
Moles of HCN produced from O2 = (Moles of O2) / 3
Moles of HCN produced from CH4 = (Moles of CH4) / 2
The mass of HCN produced can be calculated by multiplying the moles of HCN by its molar mass:
Mass of HCN = (Moles of HCN) × (Molar mass of HCN)
Similarly, calculate the mass of H2O produced using the stoichiometric ratio:
Moles of H2O produced from NH3 = (Moles of NH3) × 6
Moles of H2O produced from O2 = (Moles of O2) × 6
Moles of H2O produced from CH4 = (Moles of CH4) × 6
Mass of H2O = (Moles of H2O) × (Molar mass of H2O)
Since 100% yield is assumed, the mass of HCN and H2O calculated are the actual amounts that will be produced.