90 g of water is added to potassium oxide. The product formed is 1.120 kg of potassium hydroxide. Calculate the mass of potassium oxide used
a. 88.88g of potassium oxide is used
b.1030 g of potassium oxide is used
c. 1210 g of potassium oxide is used
d. 91.12 g of potassium oxide is used
plz
Answers
Answer:1030g
Explanation: Here, water is a limiting reagent in the reaction. We need to use the stoichiometry to find the number of moles of K2O.
Moles of KOH = 1120/56
=20moles of KOH
Mole ratio, K2O : 2KOH
Moles of K2O = 1/2× 20 moles
= 10moles of K2O
Mass of K2O = 10 ×94
=940g
But remember, K2O was in excess and therefore we add the mass of water,90g, in order to find the total mass of K2O used.
90g+940g
=1030g
90g of water is added to potassium oxide. the product formed is 1.120 kg of potassium hydroxide.
To find : The mass of potassium oxide used.
solution : According to Lavoisier's law of conservation of mass,
for any chemical balanced equation, the mass of reactants is equal to the mass of products.
here balanced chemical reaction is ...
H₂O + K₂O ⇒2KOH
the mass of H₂O and K₂O = mass of KOH
⇒mass of H₂O + mass of K₂O = mass of KOH
⇒90g + mass of K₂O = 1120g
⇒mass of K₂O = 1120 - 90 = 1030g
Therefore the mass of potassium oxide used in reaction is 1030g .