Electrolysis of 10.0 g of a binary metal chloride deposits 6.207 g of the pure metal. what is the metal?

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:

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:

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 , the by mass percent of copper is:

Whereas accounts for the number of atoms of copper in such compound, accounts for the copper's atomic mass and accounts for the copper (I) chloride's molar mass which is 70.9 g/mol, thus:

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 as shown below:

Therefore, the metal is thorium.

Best regards.