Question

If 120g of propane (c3h6) is burnt in excess of oxygen, how many grams of water are formed?

Answer 1

Answer:

194 g

194 g of water is produced if 120 gram of propane is burnt in excess of oxygen. Explanation: According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number of particles.

Answer 2

Answer:

195.84 g of water is formed.

Explanation:

The reaction will proceed as follows,

$C_3H_8+5O_2\rightarrow3CO_2+4H_2O$         (1)

Where,

C₃H₈=Propane

O₂=Oxygen

CO₂=carbon dioxide

H₂O=water

From the question we have,

The given mass of the propane=120g

The molar mass of propane=44g

The number of moles of propane (n) is,

$n=\frac{120}{44} =2.72$       (2)

From equation (1) we can see that one mole of propane reacts with four moles of oxygen.

$n_2=2.72\times 4=10.88$       (3)

n₂=number of moles of water

So, 2.72 moles of propane will react with 10.88 moles of oxygen.

$10.88=\frac{m}{M}$        (4)

Where,

m=mass of water that forms in the reaction

M=total molar mass of water=18g    (∵1 molar mass of H₂O=18g)

By substituting the values in equation (4) we get;

$10.88=\frac{m}{18}$

$m=10.88\times 18$

$m=195.84g$

Hence, 195.84 g of water is formed.