Question

Question #18399

Answer 1

This question is based on Stoichiometry, in order to answer this question we should know the chemical equation showing the reaction between the hydroxides and carbon dioxide.

2 LiOH (s) + C`O_2`(g) -------> `Li_2` `CO_3`(s) + `H_2`O(aq)

Molar mass of the following compound is

LiOH = 7+ 1 +16 = 24 g/mole

C`O_2` = 12 + 16 + 16 = 44 g/mol

2 moles of LiOH reacts with one mole of Carbon dioxide ,C`O_2`.

2 mole x 24 g/ mol of LIOH consumes 1mole x 44 g / mol of C`O_2`.

48 g of LiOH consumes 44g of C`O_2`.

1g ofLiOH consumes 44g / 48 = 0.916 g of C`O_2`

1 g of LiOH will use or remove 0.916 g of C`O_2`.

Mg`(OH)_2` ( s) + C`O_2` (g) -------> Mg `CO_3`(s) + `H_2`O (aq)

Molar mass of the following compound is

Mg`(OH)_2` = 58.3 g/mole

C`O_2` = 12 + 16 + 16 = 44 g/mol

1 mole of Mg`(OH)_2` reacts with one mole of Carbon dioxide ,C`O_2`.

1 mole x 58.3 g/ mol of Mg`(OH)_2` consumes 1 mole x 44 g / mol of C`O_2`.

58.3 g of Mg`(OH)_2` consumes 44g of C`O_2`.

1g of Mg`(OH)_2` consumes 44g / 58.3 = 0.75 g of C`O_2`

1 g of Mg`(OH)_2` will use 0.75g of C`O_2`.

2 Al`(OH)_3` (s) + 3 C`O_2`(g) ------> Al2`(CO3)3`(s) + 3 `H_2`O(aq)

Molar mass of the following compound is

Al`(OH)_3` = 78.0 g/mole

C`O_2` = 12 + 16 + 16 = 44 g/mol

2 mole of Al`(OH)_3` reacts with 3 moles of Carbon dioxide ,C`O_2`.

2 mole x 78.0 g/ mol of Al`(OH)_3` consumes 3 mole x 44 g / mol of C`O_2`.

156 g of Al`(OH)_3` consumes 132 g of C`O_2`

1g of Al`(OH)_3` consumes 132g / 156 = 0.85 g of C`O_2`

1 g of Al`(OH)_3` will use 0.85 g of C`O_2`.