Question

The combustion of propane (C 3H 8) produces CO 2 and H 2O: C 3H 8 (g) 5O2 (g) 3CO2 (g) 4 H 2O (g) The reaction of 2.5 mol of O 2 with 4.6 mol of C 3H 8 will produce ________ mol of H 2O.

Answer 1

Answer : The number of moles of $H_2O$ produced will be, 2 moles

Explanation : Given,

Moles of $O_2$ = 2.5 mol

Moles of $C_3H_8$ = 4.6 mol

First we have to calculate the limiting and excess reagent.

The given chemical reaction is,

$C_3H_8(g)+5O_2(g)\rightarrow 3CO_2(g)+4H_2O(g)$

From the balanced reaction we conclude that

As, 5 mole of $O_2$ react with 1 mole of $C_3H_8$

So, 2.5 moles of $O_2$ react with $\frac{2.5}{5}=0.5$ moles of $C_3H_8$

From this we conclude that, $C_3H_8$ is an excess reagent because the given moles are greater than the required moles and $O_2$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $H_2O$

From the reaction, we conclude that

As, 5 mole of $O_2$ react to give 4 mole of $H_2O$

So, 2.5 moles of $O_2$ react to give $\frac{2.5}{5}\times 4=2$ moles of $H_2O$

Thus, the number of moles of $H_2O$ produced will be, 2 moles