Calculate the number of grams Fe2O3 produced from 8.93 g 02.
4 Fe +302 → 2 Fe2O3

Hey there!:

Molar mass :

Fe₂O₃ = 159.69 g/mol

O₂ = 31.99 g/mol

By reaction stochiometry  :

4 Fe + 3 O₂ = 2 Fe₂O₃

3 * (31.99 g O₂ )    2 * ( 159.69 ) g Fe₂O₃

8.93 g O₂ ( mass O₂)

mass O₂ =  ( 8.93 * 2 * 159.69 ) /  ( 3 * 31.99 )

mass O₂ = 2852.0634 / 95.97

mass O₂ = 29.71 g Of Fe₂O₃

Hope this helps!

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

a electronegativity decreases beacuse the size of nucleus is bigger.