AQA Chemistry Chapter 2 Practice Questions HELP

This is quite specific but hopefully someone can help me! I’ve just done the practice questions at the end of Chapter 2 in the textbook (Oxford university press) and there is a question that even with the mark scheme, I have no idea where the answer comes from. If anyone has the same book and has done the questions and understands I would really like to know how to work out the answer to question 2)b and c. Thanks!

You're more likely to get a helpful response if you post the full question and what you've got so far

It is quite a wordy question so typing out is a pain 😩 but I have added a link to a photo of the question and my working so hopefully that will help!

Determine mol of ammonia = 3,000g/17 = 176.5 mol
This makes same mol of NO
Then in equation 2: mol NO = mol NO2 = 176.5 mol
Mass of NO2 = Mr x mol = 46 x 176.5 = 8,117.6 g
BUT there is only 80% efficiency so = 6,494g

Why do we only find the Mr of NH3 instead of 4NH3?

The 4 in front of the NH3 represents part of the molar ratio – it tells you how many moles (or molecules, or bits, or whatever) of ammonia there are in relation to the other components of the equation. Here, in reaction 1, for every 4 mol. of NH3, there are 5 of O2.

When you are calculating the actual moles of something, you'll either use mass = Mr x moles OR use the molar ratio if you already have the moles of something else (note that for things in equilibrium this may not be the case for things on opposite sides of the arrows). This means that, for example, one mole of oxygen (O2) will have a mass of about 31.998 g (mass = Mr x moles = (15.999 x 2) x 1) regardless of what the reaction is!! The big number in front of a compound in an equation does not represent the actual number of moles.

To calculate the moles of ammonia, you use the mass of NH3 (which you're given) and the Mr of NH3. NH3 represents a molecule of ammonia, and in the reaction many molecules of ammonia react. 4NH3 only represents the ratio.