Consider the following chemical equation. how many moles of seоg can be produced from 35
g of oxygen and excess selenium?
2se +302 - 2so3

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

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g