Let's go through this problem step-by-step:
1. Convert grams of Mg to moles of Mg:
- Given: [tex]\(48\)[/tex] grams of [tex]\(Mg\)[/tex].
- Molar mass of magnesium ([tex]\(Mg\)[/tex]) is [tex]\(24 \text{ g/mol}\)[/tex].
[tex]\[
\text{Moles of } Mg = \frac{\text{grams of } Mg}{\text{molar mass of } Mg} = \frac{48 \text{ g}}{24 \text{ g/mol}} = 2 \text{ mol}
\][/tex]
2. Use the stoichiometry of the reaction:
- The balanced chemical equation is:
[tex]\[
Mg + 2AgNO_3 \rightarrow 2Ag + Mg(NO_3)_2
\][/tex]
- According to the equation, [tex]\(1 \text{ mole of } Mg\)[/tex] produces [tex]\(2 \text{ moles of } Ag\)[/tex].
[tex]\[
\text{Moles of } Ag = 2 \text{ moles of } Mg \times 2 = 4 \text{ moles of } Ag
\][/tex]
3. Convert moles of Ag to grams of Ag:
- Molar mass of silver ([tex]\(Ag\)[/tex]) is [tex]\(108 \text{ g/mol}\)[/tex].
[tex]\[
\text{Grams of } Ag = \text{moles of } Ag \times \text{molar mass of } Ag = 4 \text{ moles} \times 108 \text{ g/mol} = 432 \text{ g}
\][/tex]
4. Fill in the green boxes in the conversion grid:
[tex]\[
\begin{array}{l}
\begin{array}{l|l|l|l}
48 \text{ g Mg} & 1 \text{ mol Mg} & 2 \text{ mol Ag} & 432 \text{ g Ag} \\
\hline & 24 \text{ g Mg} & 1 \text{ mol Mg} & 108 \text{ g Ag}
\end{array} = 432 \text{ g Ag}
\end{array}
\][/tex]
So, the numerical result for the final product is [tex]\(432 \text{ g Ag}\)[/tex]. The conversion values matter and confirm that 48 grams of [tex]\(Mg\)[/tex] reacts to yield 432 grams of [tex]\(Ag\)[/tex].
