Empirical formula mass
Molecular fornula mass:-180g/mol
- n=Molecular formula mass/Empirical formula mass
- m=180/30
- n=6
Molecular formula:-
- n×Empirical formula
- 6(CH_2O
- C_6H_12 O_6
Answer:
I'm pretty sure it's A. BRUSH
Explanation:
If I'm wrong let ne know please
Molar mass of copper chloride is 134.45 g/mole, so the mole of 10.5 g copper chloride is 10.5/134.45 = 0.078 mole. Molar mass of aluminum is 27 g/mole, so the mole of 12.4 g aluminium is 12.4/27 = 0.46 mole. The formula of this reaction is as follows, 2Al + 3CuCl2 ⇒ 2AlCl3 + 3Cu. Thus the molar ratio of the reactants is Al:CuCl2 = 2:3. So to react with 0.078 mole of copper chloride, will need 0.078 x 2/3 = 0.052 moles of aluminum which is less than the given amount (0.46mole). Therefore, copper chloride is the limiting reactant.
