Question

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

Answer 1

co2 is correct buddy

Answer 2

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:

$MCl_x-->M+\frac{x}{2} Cl_2$

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:

$CuCl, \ CuCl_2,\ CdCl, CeCl_2,\ CeCl_3, \ CeCl_3 \ and \ ThCl_4$

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 $CuCl$, the by mass percent of copper is:

$\% Cu=\frac{n_{Cu}m_{Cu}}{M_{CuCl}} *100\%$

Whereas $n_{Cu}$ accounts for the number of atoms of copper in such compound, $m_{Cu}$ accounts for the copper's atomic mass and $M_{CuCl}$ accounts for the copper (I) chloride's molar mass which is 70.9 g/mol, thus:

$\% Cu=\frac{1*63.5}{70.9}*100\% \\\% Cu=89.6\%$

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 $ThCl_4$ as shown below:

$\% Th=\frac{1*232}{373.8}*100\% \\\% Th=62.07\%$

Therefore, the metal is thorium.

Best regards.