Question

In a precipitation reaction where 43.7 ml of aqueous 0.255 M Pb(NO₃)₂ is mixed with 28.6 ml of 0.415 M NaCl, following the equation: Pb(NO₃)₂ (aq) + 2 NaCl (aq) → PbCl₂ (s) + 2 NaNO₃ (aq), how many moles of PbCl₂ are formed?

a) 0.0061 moles
b) 0.0122 moles
c) 0.0244 moles
d) 0.0366 moles

Answer 1

To determine the moles of PbCl₂ formed in the reaction, calculate the moles of Pb(NO₃)₂ and NaCl using the given volumes and concentrations. According to the balanced equation, 1 mole of Pb(NO₃)₂ reacts with 2 moles of NaCl to form 1 mole of PbCl₂. Therefore, the number of moles of PbCl₂ formed is equal to half the number of moles of Pb(NO₃)₂.

Step-by-step explanation:

In order to determine the number of moles of PbCl₂ formed in the precipitation reaction, we need to use the given volumes and concentrations of the reactants. First, calculate the moles of Pb(NO₃)₂ by multiplying the volume (43.7 ml) by the molar concentration (0.255 M). This gives us 0.0112 moles of Pb(NO₃)₂. Next, calculate the moles of NaCl by multiplying the volume (28.6 ml) by the molar concentration (0.415 M). This gives us 0.0119 moles of NaCl.

According to the balanced equation, 1 mole of Pb(NO₃)₂ reacts with 2 moles of NaCl to form 1 mole of PbCl₂. Therefore, the number of moles of PbCl₂ formed is equal to half the number of moles of Pb(NO₃)₂, which is 0.0056 moles.