## anonymous one year ago 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

1. Rushwr

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

2. anonymous

thank you very very much.

3. Rushwr

No problem :)