Answer:
Option d. 6.95 g
Explanation:
First of all, we state the reaction:
2NaCl + Pb(NO₃)₂ → PbCl₂ + 2NaNO₃
We determine the moles of each reactant, to state the limiting
Firstly we convert volume frm mL to L
0.200 L . 0.250M = 0.05 moles of NaCl
0.200L . 0.250M = 0.05 moles of Pb(NO₃)₂
Acording to stoichiometry we know that relation is 1:2, so the limiting reagent is the NaCl.
For 1 mol of Pb(NO₃)₂ I need 2 moles of NaCl
For 0.05 moles of Pb(NO₃)₂ I would need, the double → 0.1 moles
(We only have, 0.05 moles of NaCl)
Stoichiometry to the formed product is 2:1
From 2 moles of NaCl I produce 1 mol of PbCl₂
From 0.05 moles I would produce, the half → 0.025 moles
Let's convert the moles to mass → 0.025 mol . 278.1 g / 1mol = 6.95 g
6.95 g mass of lead (II) chloride would be produced.
Considering the balanced equation of the reaction:
2 NaCl + Pb(NO3)2 → PbCl2 + 2 NaNO3
Mole of 200 mL, 0.250 M NaCl = 0.200 x 0.25
= 0.05 mole
Mole of 200 mL, 0.25 M Pb(NO3)2 = 0.2 x 0.25
= 0.05 mole
Stoichiometric mole ratio of NaCl and Pb(NO3)2 = 2:1
Thus, NaCl is the limiting reactant.
Mole ratio of NaCl to PbCl2 = 2:1
Equivalent mole of PbCl2 = 0.05/2
= 0.025 mole
Mass of 0.025 mole PbCl2 = 0.025 x 278.1
= 6.95 g
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