Question

Consider the following chemical equation. How many moles of SeОg can be produced from 35

g of oxygen and excess selenium?
2Se +302 - 2SO3

Answer 1

0.73 mole

Step-by-step 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