Question

Barium oxalate is used as a colorant to produce the green color in fireworks. Imagine that you have been assigned to prepare barium oxalate by reacting barium hydroxide octahydrate (Ba(OH)2·8H2O) with oxalic acid dihydrate (H2C2O4·2H2O). Find the number of grams of oxalic acid dihydrate that would be required to completely react with 5.3 g of barium hydroxide octahydrate. Use correct significant figures. Do not include a unit with your answer or it will be counted wrong.
H2C2O4·2H2O 126.07 g/mol
H2C2O4 90.03 g/mol
Ba(OH)2·8H2O 315.46 g/mol
Ba(OH)2 171.34 g/mol

Answer 1

Answer:

2.1

Explanation:

Calculation of moles of $Ba(OH)_2.8H_2O$

Mass of copper = 5.3 g

Molar mass of copper = 315.46 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{5.3\ g}{315.46\ g/mol}$

Moles of $Ba(OH)_2.8H_2O$ = 0.0168 moles

According to the reaction,

$Ba(OH)_2.8H_2O + H_2C_2O_4.2H_2O\rightarrow BaC_2O_4 + 12 H_2O$

1 mole of $Ba(OH)_2.8H_2O$ react with 1 mole of  $H_2C_2O_4.2H_2O$

0.0168 moles of $Ba(OH)_2.8H_2O$ react with 0.0168 moles of  $H_2C_2O_4.2H_2O$

Moles of $H_2C_2O_4.2H_2O$ = 0.0168 moles

Molar mass of $H_2C_2O_4.2H_2O$ = 126.07 g/mol

Thus,

Mass = Moles * Molar mass = 0.0168 moles * 126.07 g/mol = 2.1 g

Answer - 2.1