Answer:
0.6g
Explanation:
Given parameters:
Mass of K = 0.5g
Mass of O₂ = 0.10g
Unknown:
Mass of K₂O = ?
Solution:
To solve this problem, let us write the reaction equation first;
4K + O₂ → 2K₂O
The reaction above delineates the balanced chemical reaction.
To solve this problem, we need to know the limiting reactant. This reactant is the one that determines the amount and extent of the reaction because it is given in short supply. The other reactant is the one in excess.
Start off by find the number of moles of the reactant;
Number of moles = [tex]\frac{mass}{molar mass}[/tex]
Molar mas of K = 39g/mol
Molar mass of O₂ = 2(16) = 32g/mol
Number of moles of K = [tex]\frac{0.5}{39}[/tex] = 0.013moles
Number of moles of O₂ = [tex]\frac{0.1}{32}[/tex] = 0.031moles
From the balanced reaction;
4 moles of K reacted with 1 mole of O₂
0.013 moles of K will react with [tex]\frac{0.013}{4}[/tex] = 0.0078 moles of O₂
We see that oxygen gas is in excess. We were given 0.031moles of the gas but only require 0.0078moles of oxygen gas.
The limiting reactant is potassium.
therefore;
4 moles of K produced 2 moles of K₂O
0.013 moles of K will produce [tex]\frac{0.013 x 2}{4}[/tex] = 0.0065moles of K₂O
to find the mass of K₂O;
Mass of K₂O = number of moles x molar mass
Molar mass of K₂O = 2(39) + 16 = 94g/mol
Mass of K₂O = 0.0065 x 94 = 0.6g
