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
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Answer:
Explanation:
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In this case, since the concept of solution concentration is majorly used in terms of moles per liters of solution or molarity (M), it is also possible to represent this chemical unit by using squared brackets, [ ].
In such a way, when focused on the concentration of glucose, C6H12O6, we can use:
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