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 5.00 x 10^3 kg each of NH3, O2, and CH4 are reacted, what mass of HCN and of H2O will be produced, assuming 100% yield?
i don't understand this at all
can someone explain please
Sure! I'd be happy to help you understand this chemical equation and how to find the mass of hydrogen cyanide (HCN) and water (H2O) produced.
First, let's analyze the balanced equation:
2 NH3(g) + 3 O2(g) + 2 CH4(g) → 2 HCN(g) + 6 H2O(g)
From this equation, we can see that 2 moles of ammonia (NH3), 3 moles of oxygen (O2), and 2 moles of methane (CH4) react to produce 2 moles of hydrogen cyanide (HCN) and 6 moles of water (H2O).
To solve the problem, we need to convert the given mass of each reactant (NH3, O2, CH4) into moles, and then use the mole ratios in the balanced equation to determine the moles of products produced. Finally, we can convert the moles of products back into mass.
Step 1: Convert the given mass of each reactant into moles.
To do this, we need to know the molar mass of each reactant. The molar mass of NH3 is approximately 17.031 g/mol, the molar mass of O2 is approximately 31.998 g/mol, and the molar mass of CH4 is approximately 16.043 g/mol.
Using these molar masses, we can calculate the number of moles for each reactant:
Number of moles of NH3 = mass of NH3 / molar mass of NH3
Number of moles of O2 = mass of O2 / molar mass of O2
Number of moles of CH4 = mass of CH4 / molar mass of CH4
Step 2: Determine the moles of products produced using the mole ratios in the balanced equation.
From the balanced equation, we know that the mole ratio of NH3 to HCN is 2:2, the ratio of O2 to HCN is 3:2, and the ratio of CH4 to HCN is 2:2.
Using these ratios, we can calculate the moles of HCN and H2O produced:
Moles of HCN = (moles of NH3 / 2) * 2
Moles of H2O = (moles of NH3 / 2) * 6
Step 3: Convert the moles of products back into mass.
Using the molar masses of HCN (approximately 27.025 g/mol) and H2O (approximately 18.015 g/mol), we can now calculate the masses of HCN and H2O:
Mass of HCN = moles of HCN * molar mass of HCN
Mass of H2O = moles of H2O * molar mass of H2O
By following these steps and plugging in the given mass values, you should be able to calculate the mass of HCN and H2O produced.
I hope this explanation helps! Let me know if you have any further questions.
Of course! Let's break down the equation and calculate the mass of HCN and H2O produced.
The balanced chemical equation you provided is:
2 NH3(g) + 3 O2(g) + 2 CH4(g) => 2 HCN(g) + 6 H2O(g)
According to the equation, it shows that 2 moles of NH3 react with 3 moles of O2 and 2 moles of CH4 to produce 2 moles of HCN and 6 moles of H2O.
To solve this problem, we need to convert the given mass of each reactant into moles, determine the limiting reactant, and then calculate the mass of products produced.
First, let's calculate the moles of each reactant:
1. NH3:
Given mass = 5.00 x 10^3 kg
Molar mass of NH3 = 17.03 g/mol
Convert mass to moles: (5.00 x 10^3 kg x 1000 g/kg) / 17.03 g/mol = 2.94 x 10^5 mol
2. O2:
Given mass = 5.00 x 10^3 kg
Molar mass of O2 = 32.00 g/mol
Convert mass to moles: (5.00 x 10^3 kg x 1000 g/kg) / 32.00 g/mol = 1.56 x 10^5 mol
3. CH4:
Given mass = 5.00 x 10^3 kg
Molar mass of CH4 = 16.04 g/mol
Convert mass to moles: (5.00 x 10^3 kg x 1000 g/kg) / 16.04 g/mol = 1.56 x 10^5 mol
Now let's determine the limiting reactant. The limiting reactant is the one that is completely consumed and limits the amount of product formed.
To find the limiting reactant, we compare the moles of each reactant to the stoichiometric ratio in the balanced equation. The reactant that produces a smaller number of moles of the product is the limiting reactant.
From the balanced equation: 2 moles NH3 : 3 moles O2 : 2 moles CH4
- NH3 moles = 2.94 x 10^5 mol
- O2 moles = 1.56 x 10^5 mol
- CH4 moles = 1.56 x 10^5 mol
The ratio of NH3 to HCN is 2:2, so the moles of HCN produced will be equal to the moles of NH3 present, which is 2.94 x 10^5 mol.
The ratio of O2 to HCN is 3:2, so the moles of HCN that could be produced from O2 are (1.56 x 10^5 mol * 2 mol HCN / 3 mol O2) = 1.04 x 10^5 mol.
The ratio of CH4 to HCN is 2:2, so the moles of HCN that could be produced from CH4 are (1.56 x 10^5 mol * 2 mol HCN / 2 mol CH4) = 1.56 x 10^5 mol.
We can see that the amount of HCN produced is limited by the amount of NH3, and the moles of HCN produced will be equal to the moles of NH3 present (2.94 x 10^5 mol).
Now, we can calculate the mass of HCN and H2O produced:
1. HCN:
Molar mass of HCN = 27.03 g/mol
Mass of HCN produced = moles of HCN * molar mass of HCN
= (2.94 x 10^5 mol) * (27.03 g/mol)
= 7.93 x 10^6 g or 7930 kg
2. H2O:
Molar mass of H2O = 18.02 g/mol
Moles of H2O produced = 6 moles H2O per 2 moles HCN * moles of HCN
= (6/2) * (2.94 x 10^5 mol)
= 8.82 x 10^5 mol
Mass of H2O produced = moles of H2O * molar mass of H2O
= (8.82 x 10^5 mol) * (18.02 g/mol)
= 1.59 x 10^7 g or 1.59 x 10^4 kg
Therefore, assuming 100% yield, 7930 kg of HCN and 1.59 x 10^4 kg of H2O will be produced.
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 5.35 103 kg each of NH3, O2, and CH4 are reacted, what mass of HCN and of H2O will be produced, assuming 100% yield?
HCN g
H2O g
Convert 5,000 kg NH3 to moles.
Convert 5,000 kg O2 to moles.
Convert 5,000 kg CH4 to moles.
Now convert each of those moles, individually, to moles of the product (HCN). The reactant producing the smallest number of moles HCN will be the limiting reagent. Then convert that many moles HCN to grams. g = moles x molar mass. Post your work if you get stuck.