Nitric acid is produced commercially by the Ostwald process, represented by the following equations.
4 NH3(g) + 5 O2(g) 4 NO(g) + 6 H2O(g)
2 NO(g) + O2(g) 2 NO2(g)
3 NO2(g) + H2O(l) 2 HNO3(aq) + NO(g)
What mass in kg of NH3 must be used to produce 4.4 multiplied by 106 kg HNO3 by the Ostwald process, assuming 100% yield in each reaction?
where do i even begin?!
Well, first of all, don't panic! I'm here to help you with a joke or two along the way.
To find the mass of NH3 needed, we need to calculate its molar mass and use the stoichiometry of the reactions.
The molar mass of NH3 (ammonia) is approximately 17 g/mol.
Now, let's start by calculating the moles of HNO3 needed. We know that 1 mole of HNO3 is approximately 63 g.
4.4 x 10^6 kg HNO3 * (1000 g/1 kg) * (1 mol/63 g) = X moles of HNO3
Now, let's use the stoichiometry of the reactions to find the moles of NH3 needed.
From the first reaction:
4 moles of NH3 react to produce 4 moles of NO.
From the third reaction:
4 moles of NO react to produce 8 moles of HNO3.
Therefore, the ratio of moles NH3 to moles HNO3 is 4:8, which simplifies to 1:2.
So, X moles of HNO3 * (1 mol NH3/2 mol HNO3) = Y moles of NH3
Finally, let's calculate the mass of NH3 needed:
Y moles of NH3 * (17 g/1 mol) * (1 kg/1000 g) = Z kg of NH3
And there you have it! Z is the mass of NH3 in kilograms needed to produce the desired amount of HNO3.
Now, here's a joke for you: Why did the scarecrow win an award? Because he was outstanding in his field!
To solve this problem, you need to use stoichiometry to calculate the mass of NH3 needed to produce a given amount of HNO3. Here's how you can approach this step-by-step:
Step 1: Write the balanced chemical equation for the conversion of NH3 to HNO3.
4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g)
3 NO2(g) + H2O(l) → 2 HNO3(aq) + NO(g)
Step 2: Determine the molar masses of NH3 and HNO3.
Molar mass of NH3 = 14.01 g/mol
Molar mass of HNO3 = 63.01 g/mol
Step 3: Calculate the number of moles of HNO3 needed.
Given mass of HNO3 = 4.4 x 10^6 kg
Convert kg to g: 4.4 x 10^6 kg = 4.4 x 10^9 g
Number of moles of HNO3 = (mass of HNO3) / (molar mass of HNO3)
= (4.4 x 10^9 g) / (63.01 g/mol)
Step 4: Use the stoichiometry to find the number of moles of NH3 required.
From the balanced equation, it can be seen that 4 moles of NH3 are required to produce 4 moles of HNO3.
Number of moles of NH3 = (number of moles of HNO3) × (4 moles of NH3 / 4 moles of HNO3)
Step 5: Finally, calculate the mass of NH3 needed.
Mass of NH3 = (number of moles of NH3) × (molar mass of NH3)
By following these steps, you should be able to determine the mass of NH3 needed to produce 4.4 x 10^6 kg of HNO3. Remember to use proper unit conversions along the way.
To solve this problem, you can use stoichiometry, which is the calculation of quantities in chemical reactions. Here's how you can approach it step-by-step:
1. Write down the balanced chemical equation for the reaction you want to use to solve the problem. In this case, it's the first equation: 4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g).
2. Determine the molar mass of NH3. The molar mass of NH3 is calculated by adding up the atomic masses of its components: nitrogen (N) and hydrogen (H). N has an atomic mass of approximately 14 g/mol, and H has an atomic mass of approximately 1 g/mol. Therefore, the molar mass of NH3 is 14 g/mol + (3 x 1 g/mol) = 17 g/mol.
3. Convert the given mass of HNO3 to moles using its molar mass. The molar mass of HNO3 is approximately 63 g/mol, so to convert kg to moles, divide the mass of HNO3 by its molar mass: 4.4 x 10^6 kg / 63 g/mol = 6.98 x 10^4 mol.
4. Use the stoichiometric ratios from the balanced equation to calculate the required amount of NH3 in moles. From the balanced equation, you can see that 4 moles of NH3 are required to produce 4 moles of NO. Therefore, the ratio is 4 moles NH3 : 4 moles NO. Since the balanced equation states that 4 moles of NH3 react with 4 moles of NO, you can conclude that 6.98 x 10^4 mol of NH3 are required to produce the same amount of NO.
5. Convert the moles of NH3 to mass using its molar mass. The molar mass of NH3 is 17 g/mol, so to convert moles to mass, multiply the moles of NH3 by its molar mass: 6.98 x 10^4 mol x 17 g/mol = 1.19 x 10^6 g.
6. Convert the mass to kilograms. Divide the calculated mass by 1000 to convert grams to kilograms: 1.19 x 10^6 g / 1000 = 1190 kg.
Therefore, approximately 1190 kg of NH3 must be used to produce 4.4 x 10^6 kg of HNO3 by the Ostwald process, assuming a 100% yield in each reaction.