Respuesta :
Answer:
0.73 mole
Explanation:
Data Given
mass of O₂= 35 g
moles of SeО₃ = ?
Reaction Given:
2Se + 3O₂ --------> 2SeO₃
Solution:
First find the mass of SeO₃ from the reaction that it produced from how much mass of oxygen (O₂).
Look at the balanced reaction
2Se + 3O₂ --------> 2SeO₃
3 mol 2 mol
So 3 mole of O₂ Produce 2 moles of SeO₃
Now
convert the moles into mass for which we have to know molar mass of SeO₃ and O₂
Molar mass of SeO₃
Molar mass of SeO₃ = 79 + 3(16) = 79 + 48
Molar mass of SeO₃ = 127 g/mol
mass of SeO₃
mass in grams = no. of moles x molar mass
mass of SeO₃ = 2 mol x 127 g/mol
mass of SeO₃ = 254 g
Molar mass of O₂ = 2(16)
Molar mass of O₂= 32 g/mol
mass of O₂
mass in grams = no. of moles x molar mass
mass of O₂ = 3 mol x 32 g/mol
mass of O₂ = 96 g
So,
96 g of O₂ Produce 254 g of SeO₃, then how many grams of SeO₃ will produced from 35.0 grams of O2
Apply unity Formula
96 g of O₂ ≅ 254 g of SeO₃
35 g of O₂ ≅ X g of SeO₃
By cross multiplication
X g of SeO₃ = 254 g x 35 g / 96 g
X g of SeO₃ = 92.6 g
92.6 g of SeO₃ will produce from 35 g of O₂
Now,
we will convert mass into moles
Formula used
no. of mole = mass in g / molar mass
Put values in above equation
no. of mole = 92.6 / 127 g/mol
no. of mole SeO₃ = 0.73 mol
SeO₃ = 0.73 mole
