Question

Determine the volume of CO₂ formed at STP if 34.2g of Aluminum tetraoxosulphate (vi) reacted with Sodium bicarbonate

Answer 1

The volume of CO₂ formed at STP if 34.2 g of aluminum tetraoxosulphate (vi) reacted with sodium bicarbonate is 13.44 L

How to calculate the volume of CO₂ formed at STP?

We shall begin by obtain the mole of CO₂ formed from the reaction. This is shown below:

• Mass of aluminum tetraoxosulphate,
  `Al_2(SO_4)_3` = 34.2 g
• Molar mass of aluminum tetraoxosulphate,
  `Al_2(SO_4)_3` = 342 g/mol
• Mole of aluminum tetraoxosulphate,
  `Al_2(SO_4)_3` = 34.2 / 342 = 0.1 mole
• Mole of CO₂ =?

`Al_2(SO_4)_3\ +\ 6NaHCO_3\ \rightarrow\ 2Al(OH)_3\ +\ 3Na_2SO_4\ +\ 6CO_2`

From the equation above,

1 mole of
`Al_2(SO_4)_3` reacted to form 6 moles of CO₂

Therefore,

0.1 mole of
`Al_2(SO_4)_3` will react to form = 0.1 × 6 = 0.6 mole of CO₂

Finally, we shall calculate the volume of CO₂ formed at STP. Details below:

• Mole of CO₂ formed = 0.6 mole
• Molar volume = 22.4 L
• Volume of CO₂ formed = ?

Volume of CO₂ formed = Mole × Molar volume

= 0.6 × 22.4

= 13.44 L