Question

OR, By heating 32.1g ammonium chloride with calcium hydroxide 10-2g NH
33-3g CaCl, and 10-8g H,O are obtained. How many gram of calcium hydroxide
takes part in the reaction? How many mole of NH,, and how many litre of NH, at
STP are formed in the reaction? (N=14, H=1)4.3 Which quantities remain fixed in​

Answer 1

10-2g of caoh is involved

1 mole=6.023×10^23

15×6.023×10^23

I think it is right

Answer 2

your answer is 75%

The first thing that you need to do here is to calculate the theoretical yield of the reaction, i.e. what you get if the reaction has a

100% yield.

the balanced equation is :-

N2+3H2➡2NH3

tells you that every 1 mole of nitrogen gas that takes part in the reaction will consume 3 moles of hydrogen gas and produce 1 mole of ammonia.

In your case, you know that 1 mole of nitrogen gas reacts with 1 mole of hydrogen gas. Since you don't have enough hydrogen gas to ensure that all the moles of nitrogen gas can react.

you can say that hydrogen gas will act as a limiting reagent, i.e. it will be completely consumed before all the moles of nitrogen gas will get the chance to take part in the reaction

So, the reaction will consume 1 mole of hydrogen gas and produce

[tex]\times{1 \ mole \ of \ hydrogen}{\divide{2 \ mole \ of \ Ammonia}{3 \ mole \ of \ Hydrogen}}[\tex] => 0.667 mole of Ammonia.