Step-by-step 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