Question

How many litres of oxygen will be required to burn 2.2 g of prapane at STP

Answer 1

The balanced chemical equation represting the reaction is

c3+h8+5o2 ➡️ 3co2+4h2o

1mole= 5❌ 22.4L

44g is stp

stoichiometry indicates that

1mole (44g) propane

=5❌22.4L o2at stp

2.2g propane = 5❌22.4❌2.2/44

5.6 litre o2 at stp.

Answer 2

The volume of oxygen gas required is 5.58 L

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of propane = 2.2 g

Molar mass of propane = 44.1 g/mol

Putting values in above equation, we get:

$\text{Moles of propane}=\frac{2.2g}{44.1g/mol}=0.0498mol$

The chemical equation for the combustion of propane follows:

$C_3H_8+5O_2\rightarrow 3CO_2+4H_2O$

By Stoichiometry of the reaction:

1 mole of propane reacts with 5 moles of oxygen gas

So, 0.0498 moles of propane will react with = $\frac{5}{1}\times 0.0498=0.249mol$ of oxygen gas

At STP:

1 mole of a gas occupies 22.4 L of volume

So, 0.249 moles of oxygen gas will occupy $\frac{22.4}{1}\times 0.249=5.58L$ of volume

Learn more about stoichiometry:

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