How many kilograms of NH3 are needed to produce 1.00×10*kg of (NH4)2SO4?

10*=? There is not enough information for me to solve, but you can use this formula to solve.

mass(1.00x10*kg) / molar mass of (NH4)2SO4 x molar mass of NH3

Almost but not quite there.

First, I assume that * meant a typo and the question was for 1.00E8 kg (NH4)2SO4.
So 1E8 kg x (2*molar mass NH3)/molar mass (NH4)2SO4

That 2 is necessary because there are two NH3 molecules in (NH4)2SO4. Or to put it another way, the equation for the formation of (NH4)2SO4 is
H2SO4 + 2NH3 ==> (NH4)2SO4

The chemical factor which Ryan and I used above is now "old fashioned" and is not taught in chemistry classes today. Instead they do this.
mols (NH4)2SO4 = grams/molar mass = ?
Now convert mols (NH4)2SO4 to mols NH3. That's 2* mols (NH4)2SO4.
Now convert mols NH3 to grams NH3. grams = mols x molar mass.
But note how much faster the use of a chemical factor is than the "new & improved" way.

To find out how many kilograms of NH3 are needed to produce 1.00×10*kg of (NH4)2SO4, we need to consider the balanced chemical equation for the reaction.

The balanced equation for the reaction between NH3 (Ammonia) and (NH4)2SO4 (Ammonium sulfate) is:

2 NH3 + H2SO4 → (NH4)2SO4

From the balanced equation, we can see that 2 moles of NH3 are required to produce 1 mole of (NH4)2SO4.

First, calculate the molar mass of (NH4)2SO4:
(NH4)2SO4 = 2(N) + 8(H) + S + 4(O) = 2(14.01 g/mol) + 8(1.01 g/mol) + 32.07 g/mol + 4(16.00 g/mol) = 132.14 g/mol

Now, let's calculate the number of moles of (NH4)2SO4:
Number of moles = mass / molar mass = (1.00×10*kg) / (132.14 g/mol) = (1.00×10*kg) * (1000 g/kg) / (132.14 g/mol) = 7.57×10* mol

Since the stoichiometry is 2:2 between NH3 and (NH4)2SO4, we need the same number of moles of NH3.

Therefore, the amount of NH3 required in kilograms is:
Mass of NH3 = Number of moles * Molar mass = (7.57×10* mol) * (17.03 g/mol) / (1000 g/kg) = 0.129 kg

So, approximately 0.129 kilograms of NH3 are needed to produce 1.00×10* kilograms of (NH4)2SO4.

To find out how many kilograms of NH3 are needed to produce a certain amount of (NH4)2SO4, you need to know the balanced chemical equation for the reaction and the molar mass of NH3.

The balanced chemical equation for the reaction between NH3 and (NH4)2SO4 can be written as:

2 NH3 + (NH4)2SO4 -> (NH4)2SO4

From this equation, we can see that 2 moles of NH3 react with 1 mole of (NH4)2SO4 to produce 1 mole of (NH4)2SO4.

First, calculate the molar mass of NH3:
- Molar mass of N = 14.01 g/mol
- Molar mass of H = 1.01 g/mol
- Multiply the molar mass of N by the number of N atoms in NH3 (1) and add it to the molar mass of H multiplied by the number of H atoms in NH3 (3):
Molar mass of NH3 = (14.01 g/mol) + (1.01 g/mol * 3) = 17.03 g/mol

Now, let's assume X kilograms of NH3 are needed to produce 1.00*10^kg of (NH4)2SO4.

Using the molar mass and the balanced chemical equation, we can set up a proportion:

2 moles of NH3 / 1 mole of (NH4)2SO4 = X kilograms of NH3 / 1.00×10^kg of (NH4)2SO4

To find X, we can rearrange the equation:

X = (2 moles of NH3 / 1 mole of (NH4)2SO4) * (1.00×10^kg of (NH4)2SO4)

Now, plug in the values:

X = (2 * 1.00×10^kg of (NH4)2SO4) / (1 mole of (NH4)2SO4)

Finally, calculate X using the given amount of (NH4)2SO4:

X = 2 * (1.00×10^kg of (NH4)2SO4) = 2.00×10^kg

Therefore, you would need 2.00×10^kg of NH3 to produce 1.00×10^kg of (NH4)2SO4.