(c) (i) Ammonium nitrate is another salt of ammonia that can be used as a fertiliser. It can be made from the reaction of ammonia with nitric acid. The balanced equation for the reaction is NH_(3)+HNO_(3)arrow NH_(4)NO_(3) A manufacturer reacts 500g of ammonia with 1000g of nitric acid. Determine the limiting reagent in the reaction. (relative formula masses NH_(3)=17.0,HNO_(3)=63.0) Show your working limiting reagent= __ (ii) A student wants to carry out the same reaction in the laboratory on a smaller scale. The student uses 50cm^3 of a 0.3moldm^-3 solution of ammonia. Calculate the mass of ammonia in the solution in grams. (relative formula mass NH_(3)=17.0) Show your working

Question

(c) (i) Ammonium nitrate is another salt of ammonia that can be used as a fertiliser. It can be made from the reaction of ammonia with nitric acid. The balanced equation for the reaction is NH_(3)+HNO_(3)arrow NH_(4)NO_(3) A manufacturer reacts 500g of ammonia with 1000g of nitric acid. Determine the limiting reagent in the reaction. (relative formula masses NH_(3)=17.0,HNO_(3)=63.0) Show your working limiting reagent= __ (ii) A student wants to carry out the same reaction in the laboratory on a smaller scale. The student uses 50cm^3 of a 0.3moldm^-3 solution of ammonia. Calculate the mass of ammonia in the solution in grams. (relative formula mass NH_(3)=17.0) Show your working

Answer 1

(c)(i) Based on our calculations from Step 1

forms approximately 29.4 mol and

forms approximately 15.9 mol, So, based on the ratio established in the balance equation,ottfried Paul

will be consumed before

. Hence, the limiting reagent =

(ii) From the calculations performed during Step 3: for 0.05*0.3 = 0.015 mol. Therefore, the mass of the ammonia in 50 cm^3 of 0.03 M solution is given by: 0.015 mol * 17.0 gm/mol (molar mass of ammonia) = 0.255 grams.

Answer 2

## Step 1:First, let's find out the moles of the given masses of ammonia (

) and nitric acid (

). We know that, ### **Number of moles =

** So, for

: Number of moles =

And for

: Number of moles =

## Step 2:The molar ratio between

and

in the balanced equation is 1:1. From the above step, if the amount of

moles is lesser than the

moles, then

becomes the limiting factor i.e.,

is the limiting reagent.## Step 3:Next, for the student’s laboratory practical, let’s find mass of ammonia in 0.3 mol dm-3 solution that is housed in a 50 cm^3 container.This problem can be solved using large and small scales. On a large scale, 1 dm^3 = 0.3 mol. On a small scale, 0.05 dm^3 (since 1 dm^3 = 1000 cm^3, 50 cm^3 = 0.05 dm^3) will yield (0.05*0.3) moles of

. Now using this information, calculate the mass using the number of moles, multiplied by the molar mass of

:### **Mass = Number of moles * Molar mass.**

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