Explanation:
3Ca(s) + 2AlCl3(aq) -> 3CaCl2(aq) + 2Al(s)
According to the question, Ca is the limiting reactant.
Therefore, we equate Ca to Aluminium which is the product whose mass we want to find
Molar mass of Ca- 40g/mol
". ". of Al- 27g/mol
3Ca --> 2Al
3×40 --> 2×27
9.2 --> x
x = 9.2×2×27= 496.8÷120=4.14
(On my left) (On my right side)
Na: 1. Na:1
Cl:2. Cl:1
Na:1 x 2 =2
Na:1 x 2=2
Cl: 1 x 2 = 2
Answer:
2Na + Cl2 -> 2NaCl
Answer:
The answer should be 40.3044 g/mol
