2NH3+H2SO4---> (NH4)2SO4

Calculate the mass of ammonia required to produce 125g of ammonium sulfate.|:

Convert 125 g ammonium sulfate to moles. moles = grams/molar mass.

Using the coefficients in the balanced equation, convert moles (NH4)2SO4 to moles NH3.

Now convert moles NH3 to grams. g = moles x molar mass.

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To calculate the mass of ammonia required, we need to use the balanced equation and mole ratios.

According to the balanced equation:
2NH3 + H2SO4 → (NH4)2SO4

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

First, we need to find the moles of (NH4)2SO4 using the given mass. We can use the molar mass of (NH4)2SO4.

The molar mass of (NH4)2SO4:
(NH4)2SO4 = (2 x 14.01 g/mol) + (8 x 1.01 g/mol) + 32.07 g/mol + (4 x 16.00 g/mol)
(NH4)2SO4 = 132.14 g/mol

Now, we can calculate the moles of (NH4)2SO4:
Moles = Mass / Molar mass
Moles = 125 g / 132.14 g/mol
Moles ≈ 0.946 mol

Since 2 moles of NH3 react with 1 mole of (NH4)2SO4, we need half the number of moles of NH3 compared to (NH4)2SO4.

Therefore, the moles of NH3 required would be:
Moles of NH3 = 0.946 mol / 2
Moles of NH3 ≈ 0.473 mol

Finally, we can calculate the mass of NH3 using the molar mass of NH3:
Molar mass of NH3 = 14.01 g/mol

Mass of NH3 = Moles of NH3 x Molar mass of NH3
Mass of NH3 = 0.473 mol x 14.01 g/mol
Mass of NH3 ≈ 6.63 g

Therefore, the mass of ammonia required to produce 125g of ammonium sulfate is approximately 6.63 grams.

To calculate the mass of ammonia required to produce ammonium sulfate, we need to use the balanced chemical equation and convert the given mass of ammonium sulfate to the mass of ammonia.

Let's start by writing the balanced chemical equation for the reaction:

2NH3 + H2SO4 → (NH4)2SO4

According to the equation, 2 moles of ammonia react with 1 mole of sulfuric acid (H2SO4) to form 1 mole of ammonium sulfate. From the coefficients in the balanced equation, we can deduce that the molar ratio of ammonia to ammonium sulfate is 2:1.

To calculate the moles of ammonium sulfate, we divide the given mass (125g) by the molar mass of ammonium sulfate:

Molar mass of (NH4)2SO4 = (2 x 14.01g/mol) + (8 x 1.01g/mol) + 32.07g/mol + (4 x 16.00g/mol)
= 132.14g/mol

Moles of (NH4)2SO4 = Mass / Molar mass
= 125g / 132.14g/mol
= 0.946 mol (approximately)

Since the molar ratio of ammonia to ammonium sulfate is 2:1, the moles of ammonia required will be twice the moles of ammonium sulfate:

Moles of ammonia = 2 x Moles of (NH4)2SO4
= 2 x 0.946 mol
= 1.892 mol (approximately)

Now, to calculate the mass of ammonia, we can use the molar mass of ammonia (NH3), which is 17.03g/mol.

Mass of ammonia = Moles of ammonia x Molar mass of NH3
= 1.892 mol x 17.03g/mol
= 32.21g (approximately)

Therefore, approximately 32.21 grams of ammonia are required to produce 125 grams of ammonium sulfate.