Question

One of the steps in the commercial process for converting ammonia to nitric acid is the conversion of nh3 to no. 4 nh3(g) + 5 o2(g) → 4 no(g) + 6 h2o(g) in a certain experiment, 1.90 g of nh3 reacts with 2.30 g of o2. (a) which is the limiting reactant? o2 nh3 (b) how many grams of no and of h2o form? 1.73 g no 1.55 g h2o (c) how many grams of the excess reactant remain after the limiting reactant is completely consumed? g (d) show that your calculations in parts (b) and (c) are consistent with the law of conservation of mass. mass of products + excess reactant g total mass of reactants g

Answer 1

Answer:- (a) $O_2$ , (b) 1.73 g NO and 1.55 g $H_2O$ , (c) 0.932 g of $NH_3$ , (d) yes, the results are consistent with law of conservation of mass.

Solution:-  The given balanced equation is:

$4NH_3+5O_2\rightarrow 4NO+6H_2O$

From balanced equation, ammonia and oxygen react in 4:5 mole ratio means 4 moles of ammonia reacts with 5 moles of oxygen to give the products.

(a) Let's calculate the moles of the reactants on dividing the grams by molar masses as:

For ammonia, $1.90gNH_3(\frac{1mole}{17g})$  = 0.112 mole

For oxygen, $2.30gO_2(\frac{1mole}{32g})$  = 0.0719 mole

From mole ratio, we need more moles of oxygen as compared to oxygen but from calculations the moles of oxygen are less than ammonia. So, oxygen is limiting reactant.

(b) To calculate the grams of each product formed we would calculate their moles first from limiting reactant that is oxygen and then multiply by their respective molar masses to convert the moles to grams as:

For grams of NO:

$0.0719moleO_2(\frac{4moleNO}{5moleO_2})(\frac{30gNO}{1moleNO})$ = 1.73 g of NO

For grams of $H_2O$ :

$0.0719moleO_2(\frac{6moleH_2O}{5moleO_2})(\frac{18gH_2O}{1moleH_2O})$  = 1.55 g of $H_2O$

So, 1.73 grams of NO and 1.55 grams of $H_2O$ would formed.

(c) Let's calculate the moles of excess reactant that is ammonia used to react with limiting reactant, oxygen as:

$0.0719moleO_2(\frac{4moleNH_3}{5moleO_2})$

= 0.0572 mole $NH_3$

Excess moles of ammonia = 0.112 - 0.0572 = 0.0548 mole

Now let's convert these excess moles of ammonia to grams on multiplying by molar mass:

$0.0548moleNH_3(\frac{17g}{1mole})$

= 0.932 g $NH_3$

So, after reaction when all the oxygen is used then 0.932 grams of ammonia would still be remaining.

(d) mass of reactants = mass of products + remaining mass of excess reactant

Let's plug in the values in it:

mass of reactants = 1.90 g + 2.30 g = 4.2 g

mass of products + remaining mass of excess reactant = 1.73 g + 1.55 g + 0.932 g

= 4.2 g

So, the results are consistent with law of conservation of mass.