Question

If 0.034 moles of potassium nitrate were formed how many moles of lead (II) nitrate did you start

Answer 1

Answer:

0.017 mole of Pb(NO₃)₂.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

2KOH + Pb(NO₃)₂ —> 2KNO₃ + Pb(OH)₂

From the balanced equation above,

1 mole of Pb(NO₃)₂ reacted to produce 2 moles of KNO₃.

Finally, we shall determine the number of mole of Pb(NO₃)₂ required to produce 0.034 mole of KNO₃. This can be obtained as follow:

From the balanced equation above,

1 mole of Pb(NO₃)₂ reacted to produce 2 moles of KNO₃.

Therefore, Xmol of Pb(NO₃)₂ will react to produce 0.034 mole of KNO₃ i.e

Xmol of Pb(NO₃)₂ = 0.034 / 2

Xmol of Pb(NO₃)₂ = 0.017 mole.

Thus, 0.017 mole of Pb(NO₃)₂ is needed for the reaction.