Youda:
1) How many grams of lead product would theoretically be produced between $17.0 \mathrm{~g}$ potassium iodide, $\mathrm{KI}$, and $25.0 \mathrm{~g}$ of lead (II) nitrate, $\mathrm{Pb}\left(\mathrm{NO}_{3}\right)$ ?
2) How much excess reactant is left over when $17.0 \mathrm{~g}$ of potassium hydroxide $(\mathrm{KOH})$ reacts with $20.0 \mathrm{~g}$ of iron (III) nitrate $\left(\mathrm{Fe}\left(\mathrm{NO}_{3}\right),\right)$ ?

Question

Accepted Answer

∻Solution∻ ‖ 1 ⋮ **Write the balanced chemical equation**: The first step is to write the balanced chemical equation for the reaction between potassium iodide (KI) and lead (II) nitrate (Pb(NO$_3$)$_2$). The reaction produces lead iodide (PbI$_2$) and potassium nitrate (KNO$_3$). The balanced equation is: $2 ext{KI} + ext{Pb(NO}_3 ext{)_2} ightarrow 2 ext{KNO}_3 + ext{PbI}_2$. ‖ ‖ 2 ⋮ **Calculate the molar mass of reactants and product**: The molar mass of KI is 166.00 g/mol, and the molar mass of Pb(NO$_3$)$_2$ is 331.20 g/mol. The molar mass of PbI$_2$ is 461.01 g/mol. ‖ ‖ 3 ⋮ **Determine the limiting reactant**: Convert the mass of each reactant to moles to find the limiting reactant. For KI: $\frac{17.0 ext{ g}}{166.00 ext{ g/mol}} = 0.1024 ext{ mol KI}$. For Pb(NO$_3$)$_2$: $\frac{25.0 ext{ g}}{331.20 ext{ g/mol}} = 0.0755 ext{ mol Pb(NO}_3 ext{)_2}$. Since the reaction requires 2 moles of KI for every 1 mole of Pb(NO$_3$)$_2$, KI is in excess and Pb(NO$_3$)$_2$ is the limiting reactant. ‖ ‖ 4 ⋮ **Calculate the theoretical yield of PbI$_2$**: Use the moles of the limiting reactant to find the theoretical yield of PbI$_2$. Since 1 mole of Pb(NO$_3$)$_2$ produces 1 mole of PbI$_2$, the moles of PbI$_2$ produced will be equal to the moles of Pb(NO$_3$)$_2$: $0.0755 ext{ mol Pb(NO}_3 ext{)_2} imes \frac{461.01 ext{ g/mol}}{1 ext{ mol PbI}_2} = 34.81 ext{ g PbI}_2$. ‖ ∻1 Answer∻ ⚹ 34.81 g of lead iodide (PbI$_2$) ⚹ ∻Key Concept∻ ⚹ Limiting reactant determines the theoretical yield ⚹ ∻Explanation∻ ⚹ The amount of product formed in a chemical reaction is limited by the reactant that is completely consumed first, known as the limiting reactant. ⚹ ∻Solution∻ ‖ 1 ⋮ **Write the balanced chemical equation**: The reaction between potassium hydroxide (KOH) and iron (III) nitrate (Fe(NO$_3$)$_3$) produces iron (III) hydroxide (Fe(OH)$_3$) and potassium nitrate (KNO$_3$). The balanced equation is: $3 ext{KOH} + ext{Fe(NO}_3 ext{)_3} ightarrow 3 ext{KNO}_3 + ext{Fe(OH)}_3$. ‖ ‖ 2 ⋮ **Calculate the molar mass of reactants**: The molar mass of KOH is 56.11 g/mol, and the molar mass of Fe(NO$_3$)$_3$ is 241.86 g/mol. ‖ ‖ 3 ⋮ **Determine the limiting reactant**: Convert the mass of each reactant to moles to find the limiting reactant. For KOH: $\frac{17.0 ext{ g}}{56.11 ext{ g/mol}} = 0.3030 ext{ mol KOH}$. For Fe(NO$_3$)$_3$: $\frac{20.0 ext{ g}}{241.86 ext{ g/mol}} = 0.0827 ext{ mol Fe(NO}_3 ext{)_3}$. Since the reaction requires 3 moles of KOH for every 1 mole of Fe(NO$_3$)$_3$, KOH is in excess. ‖ ‖ 4 ⋮ **Calculate the excess reactant remaining**: Determine the amount of KOH that reacts with Fe(NO$_3$)$_3$. $0.0827 ext{ mol Fe(NO}_3 ext{)_3} imes \frac{3 ext{ mol KOH}}{1 ext{ mol Fe(NO}_3 ext{)_3}} = 0.2481 ext{ mol KOH}$. Subtract this from the initial moles of KOH to find the excess: $0.3030 ext{ mol KOH} - 0.2481 ext{ mol KOH} = 0.0549 ext{ mol KOH}$. Convert this back to grams: $0.0549 ext{ mol KOH} imes \frac{56.11 ext{ g/mol}}{1 ext{ mol KOH}} = 3.08 ext{ g KOH}$. ‖ ∻2 Answer∻ ⚹ 3.08 g of potassium hydroxide (KOH) ⚹ ∻Key Concept∻ ⚹ Excess reactant calculation ⚹ ∻Explanation∻ ⚹ After a reaction, the amount of excess reactant remaining can be calculated by subtracting the amount that reacted from the initial amount.◊What is the limiting reactant in the first chemistry problem involving lead production from potassium iodide and lead (II) nitrate?⍭ Generate me a similar question◊⚹

Sia