Answer : The mass of Fe and Cl in sample B is 3.341 g and 6.371 g respectively.
Explanation :
Law of definite proportion : It is defined as a given compound always contains exactly the same proportion of elements by weight.
The balanced chemical reaction will be,
\(2FeCl_3\rightarrow 2Fe+3Cl_2\)
First we have to calculate the ratio of mass of \(FeCl_3\).
\(\text{Ratio}=\frac{\text{Mass of }FeCl_3 \text{in sample A}}{\text{Mass of }FeCl_3 \text{in sample B}}\)
\(\text{Ratio}=\frac{3.785g}{1.475g}=2.566\)
Now we have to calculate the mass of Fe and Cl in sample B.
\(\text{Mass of Cl in sample B}=2.566\times \text{Mass of Cl in sample A}\)
\(\text{Mass of Cl in sample B}=2.566\times 2.483g=6.371g\)
and,
\(\text{Mass of Fe in sample B}=2.566\times \text{Mass of Fe in sample A}\)
\(\text{Mass of Fe in sample B}=2.566\times 1.302g=3.341g\)
Therefore, the mass of Fe and Cl in sample B is 3.341 g and 6.371 g respectively.
