Question

Element E reacts with oxygen to produce EO2. Identify element E if 16.5 g of it react with excess oxygen to form 26.1 g of EO2

PLEASE HELP

Answer 1

I got it as manganese
\[E +O _{2}\rightarrow EO _{2}\]
As we can see Moles of E is equal to the moles of EO2
we know moles = mass divided by molar mass
If we consider the molar mass of E as M
\[n _{E} = \frac{ 16.5 }{ M }\]
Then write for the moles of EO2
Then the molar mass of EO2 is M+32
\[n _{EO _{2}} = \frac{ 26.1 }{ (M+32) }\]
We know \[n _{E}= n _{EO _{2}}\]
so
\[\frac{ 26.1 }{ M+32 } = \frac{ 16.5 }{ }\]
so WHEN U SIMPLIFY U GET 55 for M
M is the molar mass of the element E
So that is manganese

Answer 2

thank you very very much.

Answer 3

No problem :)