<span>Fe2O3 + 3CO --> 2Fe + 3CO2
</span><span>
m(Fe2O3)=213 g
m(CO)=140 g
</span>_______________
<span>n(Fe2O3)=?
m(Fe)=?
n(Fe2O3)=?
n(CO)=?
n(CO2)=?
</span>
<span>n(Fe2O3)=m(Fe2O3) / M(Fe2O3)
n(Fe2O3)= 213 g / 159,7 gmol-1 = 1,33 mol
</span>
<span>n(CO)= m(CO) / M(CO)
n(CO)= 140 g / 28,01 gmol-1 = 4,99 mol</span>
Given
Mass of NO - 824 g
Molar mass of NO - 30.01g/mol
No of moles of NO = Given mass/Molar mass
No of moles of NO = 824/30.01= 27.45 mole
Hence 27.5 moles of NO are formed!
First take all percents and make them grams. Since you're not given a overall molar mass you can assume it is 100 and therefore the percents are their masses.
So you have 14.31g Carbon, 1.2g Hydrogen, and 84.49g of Chlorine. Next you divide each by their molar masses to get moles of each.
Carbon= <u>14.31</u>g Hydrogen= <u>1.2</u>g Chlorine= <u>85.49</u>g
12.01g 1.01g 35.45g
= 1.19moles = 1.188moles = 2.411moles
Next you divide each of those numbers by the smallest, in this case, Hydrogen.
Thus,
Carbon= <u>1.19moles</u> Hydrogen= <u>1.188moles</u> Chlorine= <u>2.411moles</u>
1.188moles 1.188moles 1.188moles
=1.002 =1 =2.02
These are all close enough to round, so your final empirical formula is: CHCl2
Hope that helps!!
