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Sagot :
SOLUTION:
Step 1: Write the balanced chemical equation.
2NaCl + Pb(NO₃)₂ → 2NaNO₃ + PbCl₂
Step 2: Calculate the molar mass of reactants and products.
The molar masses of Na, Cl, Pb, N, and O are 22.990 g, 35.45 g, 207.2 g, 14.007 g, and 15.999 g, respectively.
• For NaCl
[tex]\begin{aligned} MM_{\text{NaCl}} & = \text{22.990 g + 35.45 g} \\ & = \text{58.44 g} \end{aligned}[/tex]
• For Pb(NO₃)₂
[tex]\begin{aligned} MM_{\text{Pb}(\text{NO}_3)_2} & = \text{207.2 g + 2(14.007 g) + 6(15.999 g)} \\ & = \text{331.2 g} \end{aligned}[/tex]
• For PbCl₂
[tex]\begin{aligned} MM_{\text{PbCl}_2} & = \text{207.2 g + 2(35.45 g)} \\ & = \text{278.1 g} \end{aligned}[/tex]
Step 3: Calculate the number of moles of PbCl₂ formed by each reactant.
• Using NaCl
Based on the balanced chemical equation, 2 moles of NaCl is stoichiometrically equivalent to 1 mole of PbCl₂.
[tex]\begin{aligned} \text{moles of} \: \text{PbCl}_2 & = \text{15.3 g NaCl} \times \frac{\text{1 mol NaCl}}{\text{58.44 g NaCl}} \times \frac{\text{1 mol} \: \text{PbCl}_2}{\text{2 mol NaCl}} \\ & = \text{0.1309 mol} \end{aligned}[/tex]
• Using Pb(NO₃)₂
Based on the balanced chemical equation, 1 mole of Pb(NO₃)₂ is stoichiometrically equivalent to 1 mole of PbCl₂.
[tex]\begin{aligned} \text{moles of} \: \text{PbCl}_2 & = \text{60.8 g} \: \text{Pb}(\text{NO}_3)_2 \times \frac{\text{1 mol} \: \text{Pb}(\text{NO}_3)_2}{\text{331.2 g} \: \text{Pb}(\text{NO}_3)_2} \times \frac{\text{1 mol} \: \text{PbCl}_2}{\text{1 mol} \: \text{Pb}(\text{NO}_3)_2} \\ & = \text{0.18357 mol} \end{aligned}[/tex]
Step 4: Determine the limiting reactant.
Since NaCl produced less amount of PbCl₂ than Pb(NO₃)₂,
NaCl is the limiting reactant.
Step 5: Calculate the mass of PbCl₂ formed.
The mass of the product that can be formed is dictated by the limiting reactant. In this case, we will start at the number of moles of PbCl₂ formed from the limiting reactant (NaCl) which is equal to 0.1309 mol.
[tex]\begin{aligned} \text{mass of} \: \text{PbCl}_2 & = \text{0.1309 mol} \: \text{PbCl}_2 \times \frac{\text{278.1 g} \: \text{PbCl}_2}{\text{1 mol} \: \text{PbCl}_2} \\ & = \boxed{\text{36.4 g}} \end{aligned}[/tex]
Hence, the mass of PbCl₂ produced is 36.4 g.
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