Question

1. an unknown group iia metal oxide weighs 19.8 g and would equate to 0.491 mol. use stoichiometric calculations to identify the unknown metal.

Explanation:

Step1: Recall molar - mass formula

The molar - mass formula is $M=\frac{m}{n}$, where $M$ is molar mass, $m$ is mass, and $n$ is the number of moles.

Step2: Identify given values for metal oxide problem

For the unknown group IIA metal oxide problem, $m = 19.8\ g$ and $n=0.491\ mol$.

Step3: Calculate molar mass of metal oxide

Using the formula $M=\frac{m}{n}$, we substitute the values: $M=\frac{19.8\ g}{0.491\ mol}\approx40.33\ g/mol$.

Step4: Consider the formula of group IIA metal oxide

The general formula of a group IIA metal oxide is $MO$, where $M$ is the metal and $O$ is oxygen. The molar mass of oxygen is $M_O = 16.00\ g/mol$.

Step5: Calculate molar mass of the metal

Let the molar mass of the metal be $M_M$. Then $M = M_M+M_O$. So $M_M = M - M_O$. Substituting the values, $M_M=40.33\ g/mol - 16.00\ g/mol=24.33\ g/mol$. The metal with a molar mass close to $24.33\ g/mol$ is magnesium ($Mg$).

Answer:

The unknown metal is magnesium ($Mg$).