How many kilograms of NH3 are needed to produce 2.90 x 10^5 kg of (NH4)2SO4?
first, write the chemical reaction:
NH3 + H2SO4 -> (NH4)2SO4
then we balance this:
2NH3 + H2SO4 -> (NH4)2SO4
then we find the weight of NH3 and (NH4)2SO4. We can do this by getting each atomic mass of the element in the chemical formula, and add them. It can be found on periodic table of elements.
NH3 = 17 kg/kmol
(NH4)2SO4 = 132 kg/kmol
then, we get the number of moles of the product by dividing the given by the mass:
2.90*10^5 kg (NH4)2SO4 / 132 kg/kmol = 2196.97 kmol (NH4)2SO4
then we get the number of moles of NH3 required (stoichiometric ratios from the balanced equation):
2196.97 kmol (NH4)2SO4 * (2 kmol NH3 / 1 kmol (NH4)2SO4) = 4393.93 kmol NH3
finally, we multiply this by the mass of NH3 to get the total mass required:
4393.93 kmol NH3 * (17 kg/kmol) = 74696.97 kg NH3
hope this helps~ :)
Wow, that's a tough one! But don't worry, this is no joking matter. To calculate the amount of NH3 needed, we need to look at the stoichiometry of the reaction. In the balanced equation for the reaction, we have:
2 NH3 + H2SO4 -> (NH4)2SO4
From the equation, we can see that it takes 2 moles of NH3 to produce 1 mole of (NH4)2SO4. So, let's calculate the moles of (NH4)2SO4 first:
Moles of (NH4)2SO4 = mass / molar mass
Moles of (NH4)2SO4 = 2.90 x 10^5 kg / (132.14 g/mol + 2*14.01 g/mol + 4*1.01 g/mol)
Then we can calculate the moles of NH3 needed:
Moles of NH3 = 2 * Moles of (NH4)2SO4
Now we just need to convert moles of NH3 to kilograms:
Mass of NH3 = Moles of NH3 * molar mass of NH3
And there you have it! The amount of kilograms of NH3 needed to produce 2.90 x 10^5 kg of (NH4)2SO4.
To determine the number of kilograms of NH3 needed to produce (NH4)2SO4, we need to understand the stoichiometry of the reaction. The balanced chemical equation for the reaction is:
2 NH3 + H2SO4 → (NH4)2SO4
From the balanced equation, we can see that 2 moles of NH3 react with 1 mole of H2SO4 to produce 1 mole of (NH4)2SO4.
To find the number of moles of (NH4)2SO4, we can use the given mass and molar mass. The molar mass of (NH4)2SO4 is calculated as follows:
Molar mass of (NH4)2SO4 = (2 × molar mass of NH4) + molar mass of SO4
= (2 × (1 × atomic mass of N + 4 × atomic mass of H)) + atomic mass of S + 4 × atomic mass of O
= (2 × (1 × 14.01 + 4 × 1.01)) + 32.07 + 4 × 16.00
= 132.14 g/mol
Now, we can calculate the number of moles of (NH4)2SO4:
Moles of (NH4)2SO4 = mass of (NH4)2SO4 / molar mass of (NH4)2SO4
= 2.90 x 10^5 kg / (132.14 g/mol × 1000 g/kg)
= 2.188 x 10^3 mol
Since the stoichiometry of the reaction is 2:1 (NH3:H2SO4), the number of moles of NH3 required is half of the number of moles of (NH4)2SO4:
Moles of NH3 = 1/2 × Moles of (NH4)2SO4
= 1/2 × 2.188 x 10^3 mol
= 1.094 x 10^3 mol
Finally, we can convert the number of moles of NH3 to kilograms using the molar mass of NH3:
Mass of NH3 = Moles of NH3 × molar mass of NH3
= 1.094 x 10^3 mol × (1 × atomic mass of N + 3 × atomic mass of H)
= 1.094 x 10^3 mol × (14.01 + 3 × 1.01)
= 1.094 x 10^3 mol × 17.04 g/mol
= 18.66 x 10^3 g
= 18.66 kg
Therefore, approximately 18.66 kilograms of NH3 are needed to produce 2.90 x 10^5 kilograms of (NH4)2SO4.
To find the number of kilograms of NH3 needed to produce (NH4)2SO4, we need to look at the chemical equation for the reaction.
The chemical equation for the formation of (NH4)2SO4 from NH3 is:
2 NH3 + H2SO4 → (NH4)2SO4
From the equation, we can see that 2 moles of NH3 react to produce 1 mole of (NH4)2SO4.
To find the number of moles of (NH4)2SO4 produced, we divide the given mass of (NH4)2SO4 by its molar mass:
Molar mass of (NH4)2SO4 = (2 × molar mass of NH4) + molar mass of SO4
= (2 × 18.03 g/mol) + 96.06 g/mol
≈ 132.12 g/mol
Given mass of (NH4)2SO4 = 2.90 × 10^5 kg = 2.90 × 10^8 g
Number of moles of (NH4)2SO4 = (mass of (NH4)2SO4) / (molar mass of (NH4)2SO4)
= (2.90 × 10^8 g) / (132.12 g/mol)
≈ 2.19 × 10^6 mol
Since 2 moles of NH3 react to produce 1 mole of (NH4)2SO4, we will require half the number of moles of NH3:
Number of moles of NH3 = (1/2) × (number of moles of (NH4)2SO4)
≈ (1/2) × (2.19 × 10^6 mol)
= 1.095 × 10^6 mol
Finally, to convert moles of NH3 to kilograms, we multiply by the molar mass of NH3:
Molar mass of NH3 = 17.03 g/mol
Mass of NH3 = (number of moles of NH3) × (molar mass of NH3)
≈ (1.095 × 10^6 mol) × (17.03 g/mol)
≈ 18.67 × 10^6 g
So, approximately 18.67 × 10^6 grams (or 1.867 × 10^4 kg) of NH3 are needed to produce 2.90 × 10^5 kg of (NH4)2SO4.