how many kilograms of NH3 are needed to produce 2.90x10^5kg of (NH4)2SO4?

the balanced equation is:
2 NH3 + H2SO4 --> (NH4)2SO4

This question has already been posted and answered

To calculate the amount of NH3 needed to produce a given amount of (NH4)2SO4, we first need to determine the molar ratio between NH3 and (NH4)2SO4 based on the balanced equation.

The balanced equation shows that 2 moles of NH3 react to produce 1 mole of (NH4)2SO4.

Now, let's calculate the number of moles of (NH4)2SO4 in 2.90x10^5 kg.

First, convert the mass of (NH4)2SO4 from kilograms to grams by multiplying by 1000:
2.90x10^5 kg * 1000 g/kg = 2.90x10^8 g.

Next, determine the molar mass of (NH4)2SO4 by adding up the atomic masses of each element:
(N = 14.01 g/mol) * 2 + (H = 1.01 g/mol) * 8 + (S = 32.07 g/mol) + (O = 16.00 g/mol) * 4 = 132.14 g/mol.

Now, calculate the number of moles of (NH4)2SO4:
2.90x10^8 g / 132.14 g/mol = 2.19x10^6 mol.

Since the molar ratio of NH3 to (NH4)2SO4 is 2:1, we divide the number of moles of (NH4)2SO4 by 2 to get the number of moles of NH3:
2.19x10^6 mol / 2 = 1.095x10^6 mol.

Finally, convert the number of moles of NH3 to its mass in kilograms by multiplying by its molar mass. The molar mass of NH3 is (N = 14.01 g/mol) * 3 + (H = 1.01 g/mol) * 3 = 17.03 g/mol:
1.095x10^6 mol * 17.03 g/mol = 1.866x10^7 g.

Therefore, the amount of NH3 needed to produce 2.90x10^5 kg of (NH4)2SO4 is 1.866x10^4 kg.