5. Directions Drag and drop the correct answer choice to each answer blank. Or touch the answer choice followed by the answer blank. This balanced equation represents the reaction of Zn and Pb(NO_(3))_(2) Zn+Pb(NO_(3))_(2)arrow Zn(NO_(3))_(2)+Pb When 34 g of Zn reacts with 27 g of Pb(NO_(3))_(2) 12 g of Pb is produced. How many grams of Zn(NO_(3))_(2) are produced? Move the correct numerical answer to the first box, and the correct unit to the second box. :38 square square g/mL

To solve this problem, we need to use the concept of stoichiometry and the balanced chemical equation to determine the amount of Zn(NO3)2 produced.<br /><br />Given information:<br />- Balanced chemical equation: Zn + Pb(NO3)2 → Zn(NO3)2 + Pb<br />- Mass of Zn: 34 g<br />- Mass of Pb(NO3)2: 27 g<br />- Mass of Pb produced: 12 g<br /><br />Step 1: Calculate the molar mass of each compound.<br />Molar mass of Zn = 65.38 g/mol<br />Molar mass of Pb(NO3)2 = 331.21 g/mol<br />Molar mass of Zn(NO3)2 = 189.39 g/mol<br />Molar mass of Pb = 207.2 g/mol<br /><br />Step 2: Calculate the number of moles of each compound.<br />Moles of Zn = 34 g / 65.38 g/mol = 0.52 mol<br />Moles of Pb(NO3)2 = 27 g / 331.21 g/mol = 0.081 mol<br /><br />Step 3: Determine the limiting reactant.<br />Since the balanced equation shows that 1 mol of Zn reacts with 1 mol of Pb(NO3)2, the limiting reactant is Pb(NO3)2.<br /><br />Step 4: Calculate the amount of Zn(NO3)2 produced.<br />Moles of Zn(NO3)2 produced = Moles of Pb(NO3)2 (limiting reactant) = 0.081 mol<br />Mass of Zn(NO3)2 produced = Moles of Zn(NO3)2 × Molar mass of Zn(NO3)2<br />Mass of Zn(NO3)2 produced = 0.081 mol × 189.39 g/mol = 15.36 g<br /><br />Therefore, the correct answer is:<br />15.36 g<br />$g$