HOW MANY MOLES OF NH3 WOULD BE FORMED FROM 50 GRAMS OF (NH4)2SO4 IN THE FOLLOWING EQUATION?

(NH4)2SO4 + 2 NaOH=2NH3 + 2 H2O + Na2SO4

This is the same type as the C6H6/CO2 problem. You try it following the template I gave to you and see if you can do it. If you get stuck, let me know what you don't understand.

To find out how many moles of NH3 would be formed from 50 grams of (NH4)2SO4, you need to use the given equation and convert the mass of (NH4)2SO4 to moles.

Let's start by calculating the molar mass of (NH4)2SO4:
(NH4)2SO4 = 2(NH4) + SO4
= 2(NH4) + S + 4O
= 2(14.01 g/mol) + 32.07 g/mol + 4(16.00 g/mol)
= 28.02 g/mol + 32.07 g/mol + 64.00 g/mol
= 124.09 g/mol

Now, we can use the molar mass to calculate the number of moles of (NH4)2SO4:
moles = mass / molar mass
moles = 50 g / 124.09 g/mol
moles = 0.403 moles

From the balanced equation, we can see that each mole of (NH4)2SO4 will produce 2 moles of NH3.
So, the number of moles of NH3 formed can be calculated as:
moles of NH3 = 0.403 moles x 2
moles of NH3 = 0.806 moles

Therefore, 0.806 moles of NH3 would be formed from 50 grams of (NH4)2SO4.

To determine how many moles of NH3 would be formed from 50 grams of (NH4)2SO4, we need to use the concept of moles and molar masses.

First, let's find the molar mass of (NH4)2SO4:
(NH4)2SO4
[N: 14.01 g/mol × 2] + [H: 1.01 g/mol × 8] + [S: 32.07 g/mol] + [O: 16.00 g/mol × 4]
28.02 g/mol + 8.08 g/mol + 32.07 g/mol + 64.00 g/mol
= 132.17 g/mol

Next, we need to calculate the moles of (NH4)2SO4:
Moles = Mass / Molar mass
Moles = 50 g / 132.17 g/mol
Moles ≈ 0.378 mol

Now, let's look at the balanced equation:
(NH4)2SO4 + 2 NaOH = 2 NH3 + 2 H2O + Na2SO4

According to the balanced equation, the molar ratio between (NH4)2SO4 and NH3 is 1:2. This means that for every 1 mole of (NH4)2SO4, we will have 2 moles of NH3.

Therefore, if we have 0.378 moles of (NH4)2SO4, we can calculate the moles of NH3 as follows:
Moles of NH3 = Moles of (NH4)2SO4 × (2 moles of NH3 / 1 mole of (NH4)2SO4)
Moles of NH3 = 0.378 mol × (2 mol/1 mol)
Moles of NH3 = 0.756 mol

Hence, from 50 grams of (NH4)2SO4, approximately 0.756 moles of NH3 would be formed in the given equation.