Question

Hlo....guyz

I need ur help......

n' plz be a little fast....

Answer 1

●Burning of magnesium :-

2Mg + O2 ---> 2MgO
______

●Mass of Mg = 1 gm

●molar mass of Mg = 24 gm

●Moles of Mg = 1/24 Mol = 0.041 Mol
_______

●Mass of O2 = 0.5 gm

●Molar mass of O2 = 32

●Moles of O2 = 0.5/32 = 0.015 Mol
______-

●1 mol of O2 react with 2 mol of Mg

Therefore

●0.015 mom of O2 will react with = 2×0.015 = 0.030 mol of Mg
_____-

●We have 0.041 Mol of Mg

So

=> 0.041 - 0.030 = 0.011 mol of Mg left.

___________________

●amount of MgO formed in reaction...

●1 mol of O2 given 2 mol of MgO

●0.015 mol O2 will be 0.03 mol of MgO

●Molar mass of MgO = 40.3 gm

_____-__-

●mass of MgO formed = Molar mass × moles

= 40.3 × 0.03 = 1.209 gm
________-____-___-__-__[ANSWER]

Answer 2

Hlo...Here is your ans.