__ 9. How many grams of silver are produced when 43.5 g of strontium is added to silver nitrite? underline ( )Sr+underline ( )AgNO_(2)arrow underline ( )Sr(NO_(2))_(2)+underline ( )Ag a. 107gAg b. 151 g Ag c. 1399 Ag d. 180. g Ag __

To solve this problem, we need to use stoichiometry based on the balanced chemical equation. The balanced equation for the reaction is:<br /><br />\[ \text{Sr} + 2\text{AgNO}_2 \rightarrow \text{Sr(NO}_2)_2 + 2\text{Ag} \]<br /><br />From the balanced equation, we see that 1 mole of Sr produces 2 moles of Ag.<br /><br />First, calculate the moles of strontium (Sr) used:<br /><br />- Molar mass of Sr = 87.62 g/mol<br /><br />\[ \text{Moles of Sr} = \frac{43.5 \text{ g}}{87.62 \text{ g/mol}} \approx 0.496 \text{ mol} \]<br /><br />According to the balanced equation, 1 mole of Sr produces 2 moles of Ag. Therefore, 0.496 moles of Sr will produce:<br /><br />\[ 0.496 \text{ mol Sr} \times 2 \text{ mol Ag/mol Sr} = 0.992 \text{ mol Ag} \]<br /><br />Now, calculate the mass of silver (Ag) produced:<br /><br />- Molar mass of Ag = 107.87 g/mol<br /><br />\[ \text{Mass of Ag} = 0.992 \text{ mol} \times 107.87 \text{ g/mol} \approx 107 \text{ g} \]<br /><br />Therefore, the correct answer is:<br /><br />a. \(107 \text{ g Ag}\)