Chemistry, asked by shivanigowri272001, 8 months ago

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

Answered by JamesOwino
2

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

Answered by abhi178
2

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 .

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