Question

What mass of Oxygen reacts with 9.2g of Sodium? using this equation: 4Na + O2 = 2Na2O please show me how to do it :)

Answer 1

Hey mate,

this is a mole concept question.

Given:

mass of sodium = 9.2 gm

reaction = 4Na + O2 -------> 2Na2O

=>atomic mass of sodium = rmm = 23

thus,

the no. of moles (n) = weight/rmm = 9.2/23

=> 0.4 moles

=> atomic mass of sodium dioxide = 23+ 32 = 55

the given ratio =

4: 1 : 2

for the given ratio ,

=> sodium : sodium dioxide

=> 4 : 2

thus moles=> 0.4 : x

x = 0.2 moles

thus,

Na2O moles = 0.2

weight = rmm × moles = 0.2 × 55 = 11.0 gm

Answer 2

Answer:

$\huge\underline\bold\purple{Answer}$

4Na + O2 = 2Na2O

By the stoichiometry of this reaction if 5 mol natrium react, then 2.5 mol

Na2O should result.

Explanation:

The molecular mass of natrium oxide is

61.98

g⋅mol power-1 . If 5mol natrium react, then 5/2mol×61.98g⋅mol power-1=

154.95g natrium oxide should result.

So what have I done here? First, I had a balanced chemical equation (this is the important step; is it balanced?). Then I used the stoichiometry to get the molar quantity of product, and converted this molar quantity to mass1.