Answer:
The limiting reactant for this reaction is the HCl
Explanation:
This is my reaction:
4HCl(g)+O2(g)→2H2O(l)+2Cl2(g)
Molar mass O2 = 32 g/mol
Molar mass HCl = 36,45 g/mol
Mass / Molar mass = Moles
Moles HCl : 63,1 g / 36,45 g/m = 1,73 moles
Moles O2: 17,2 g / 32g/m = 0,54 moles
4 moles of HCl react with 1 mol of O2, according to reaction, so
1,73 moles of HCl, are going to react with, how many moles of O2.
4 moles HCl ___ 1 mol O2
1,73 moles HCl ___ (1,73 . 1)/ 4 = 0,43 moles of O2
O2 is my excess reagent because I need 0,43 moles and I have 0,54 so I have moles in excess.
1 mol of O2 are going to react with 4 moles of HCl
0,54 moles of O2 are going to react with, how many moles of HCl ?
1 mol O2 ____ 4 moles HCl
0,54 mol O2 ___ (0,54 . 4)/ 1 = 2,16 moles
I need 2,16 moles to consume my moles of O2, but I only have 1,73 moles, tha's why the HCl is mi limiting reactant.
