Answer:
Thorium
Explanation:
Hello,
In this case, we have four options for the metals to be copper (63.5g/mol), cadmium (112.4g/mol), cerium (140g/mol) and thorium (232g/mol) in addition to the expected chemical reaction:
[tex]MCl_x-->M+\frac{x}{2} Cl_2[/tex]
We must consider that 10.0 grams of the binary metal chloride yields 6.207 f of the pure metal, nonetheless, based on each metal's oxidation states we have seven options which are listed below:
[tex]CuCl, \ CuCl_2,\ CdCl, CeCl_2,\ CeCl_3, \ CeCl_3 \ and \ ThCl_4[/tex]
Now, a suitable strategy is to compute the metal's by mass percent in each option and compare it with the actual metal's by mass percent inferred from the statement which is 62.07% (6.207/10.0*100%). For instance, for [tex]CuCl[/tex], the by mass percent of copper is:
[tex]\% Cu=\frac{n_{Cu}m_{Cu}}{M_{CuCl}} *100\%[/tex]
Whereas [tex]n_{Cu}[/tex] accounts for the number of atoms of copper in such compound, [tex]m_{Cu}[/tex] accounts for the copper's atomic mass and [tex]M_{CuCl}[/tex] accounts for the copper (I) chloride's molar mass which is 70.9 g/mol, thus:
[tex]\% Cu=\frac{1*63.5}{70.9}*100\% \\\% Cu=89.6\%[/tex]
Such value does not dovetail with the percent computed from the statement (62.07%), in this manner, after doing it for all the metals, the only one that matches is the [tex]ThCl_4[/tex] as shown below:
[tex]\% Th=\frac{1*232}{373.8}*100\% \\\% Th=62.07\%[/tex]
Therefore, the metal is thorium.
Best regards.
