Question

question 7 (1 point) retake question
when 0.25 mg_(s) is added to 250ml of 3.00 mol/l hcl_(aq), hydrogen gas is produced. the magnesium becomes mg^(2+)_(aq) ions and stays in the solution. start by writing ionic and net ionic equations for the reaction.
what is the of mg^(2+) ion
mg^(2+) mol/l

Explanation:

Step1: Write the balanced chemical equation

The reaction between magnesium (Mg) and hydrochloric acid (HCl) is a single - displacement reaction. The chemical equation is:
$$\ce{Mg(s) + 2HCl(aq) -> MgCl_{2}(aq) + H_{2}(g)}$$
In ionic form, HCl dissociates into $\ce{H^{+}}$ and $\ce{Cl^{-}}$ ions, and $\ce{MgCl_{2}}$ dissociates into $\ce{Mg^{2+}}$ and $\ce{Cl^{-}}$ ions. So the ionic equation is:
$$\ce{Mg(s) + 2H^{+}(aq) + 2Cl^{-}(aq) -> Mg^{2+}(aq) + 2Cl^{-}(aq) + H_{2}(g)}$$
The net ionic equation (after canceling the spectator ions, which are $\ce{Cl^{-}}$ ions here) is:
$$\ce{Mg(s) + 2H^{+}(aq) -> Mg^{2+}(aq) + H_{2}(g)}$$

Step2: Calculate the moles of Mg and HCl

First, calculate the moles of Mg. The molar mass of Mg is approximately $24.31\space g/mol$. The mass of Mg is $0.25\space g$ (assuming the unit is grams, since $0.25\space Mg$ is likely a typo and should be $0.25\space g\space Mg$).
Moles of $\ce{Mg}$, $n_{Mg}=\frac{m}{M}=\frac{0.25\space g}{24.31\space g/mol}\approx0.0103\space mol$
Moles of $\ce{HCl}$: The volume of HCl solution is $V = 250\space ml=0.25\space L$, and the concentration $C = 3.00\space mol/L$.
Using the formula $n = C\times V$, moles of $\ce{HCl}$, $n_{HCl}=3.00\space mol/L\times0.25\space L = 0.75\space mol$

Step3: Determine the limiting reactant

From the balanced chemical equation $\ce{Mg(s) + 2HCl(aq) -> MgCl_{2}(aq) + H_{2}(g)}$, the mole ratio of $\ce{Mg}$ to $\ce{HCl}$ is $1:2$.
Moles of HCl required to react with $0.0103\space mol$ of Mg is $n_{HCl\space required}=2\times n_{Mg}=2\times0.0103\space mol = 0.0206\space mol$
We have $0.75\space mol$ of HCl, which is much more than $0.0206\space mol$. So Mg is the limiting reactant.

Step4: Calculate the moles of $\ce{Mg^{2+}}$ produced

From the balanced chemical equation, 1 mole of Mg produces 1 mole of $\ce{Mg^{2+}}$. Since Mg is the limiting reactant, moles of $\ce{Mg^{2+}}$ produced, $n_{Mg^{2+}}=n_{Mg}=0.0103\space mol$

Step5: Calculate the concentration of $\ce{Mg^{2+}}$

The volume of the solution is $0.25\space L$ (the volume of the HCl solution, and the volume change due to the addition of Mg is negligible).
Concentration of $\ce{Mg^{2+}}$, $[\ce{Mg^{2+}}]=\frac{n_{Mg^{2+}}}{V}=\frac{0.0103\space mol}{0.25\space L}\approx0.0412\space mol/L$ (If we consider the initial typo and the mass of Mg is actually $0.25\space mol$ (unlikely, but if we assume that), then moles of Mg is $0.25\space mol$. Then moles of HCl required to react with $0.25\space mol$ of Mg is $0.5\space mol$, and HCl is still in excess (since we have $0.75\space mol$ of HCl). Then moles of $\ce{Mg^{2+}}$ is $0.25\space mol$, and concentration $[\ce{Mg^{2+}}]=\frac{0.25\space mol}{0.25\space L}=1.00\space mol/L$. But the more likely case is the mass is in grams. However, if we assume that the $0.25$ is in moles (if it was a unit error and should be $0.25\space mol\space Mg$), let's recalculate:
If $n_{Mg} = 0.25\space mol$, moles of HCl required is $2\times0.25 = 0.5\space mol$, and we have $0.75\space mol$ of HCl, so Mg is still the limiting reactant (since HCl is in excess). Then moles of $\ce{Mg^{2+}}$ is $0.25\space mol$, and concentration $[\ce{Mg^{2+}}]=\frac{0.25\space mol}{0.25\space L}=1.00\space mol/L$. But given that the mass of Mg is more likely in grams, but if we take the problem as $0.25\space mol\space Mg$ (due to the possible typo), the concentration is $1.00\space mol/L$. But if we take the mass as $0.25\space g$, the concentration is approximately $0.04\space mol/L$. However, let's re - examine the problem. If the initial amount of Mg is $0.25\space mol$ (assuming the unit was wrong), then:

From the reaction, 1 mole of Mg produces 1 mole of $\ce{Mg^{2+}}$. The volume of the solution is $0.25\space L$. So $[\ce{Mg^{2+}}]=\frac{0.25\space mol}{0.25\space L}=1.0\space mol/L$. (This is a more reasonable answer if we assume that the $0.25$ is in moles, as the concentration of HCl is $3\space mol/L$ and the volume is $0.25\space L$, so HCl is in large excess, and Mg is the limiting reactant. If Mg is $0.25\space mol$, then it will produce $0.25\space mol$ of $\ce{Mg^{2+}}$ in $0.25\space L$ solution, so the concentration is $1.0\space mol/L$)

Answer:

If we assume the mass of Mg is $0.25\space mol$ (correcting the unit error), the concentration of $\ce{Mg^{2+}}$ is $\boldsymbol{1.00\space mol/L}$ (or if the mass is $0.25\space g$, approximately $0.04\space mol/L$). But based on the context of the concentration of HCl being $3\space mol/L$ and volume $0.25\space L$, a more likely intended value is when Mg is $0.25\space mol$, so the answer is $\boldsymbol{1.00\space mol/L}$ (or $1\space mol/L$)