How many moles of precipitate will be formed when 122.7 mL of 0.150 M LiBr is reacted with excess Pb(NO₃)₂ in the following chemical reaction? 2 LiBr(aq) + Pb(NO₃)₂(aq) → PbBr₂(s) + 2 LiNO₃(aq)

Answer 1

Hello, Rita,

Given the balanced equation:

2 LiBr(aq) + Pb(NO₃)₂(aq) → PbBr₂(s) + 2 LiNO₃(aq)

We see that 1 moe of PbBr₂ is formed from every 2 moles of LiBr (with excess lead nitrate). If we find the moles of LiBr, we can take half that amount to find the moles of PbBr₂ produced. We have the concentration and volume of LiBr, so let's find the numebr of moles that represents. 0.150 M LiBr means that we have 0.15 moles per liter. We can multiply the liter times the concentration to find moles. But we'll need to convert ml to Liters for the units to match. Divide 122.7 ml by 1000 to find liters (0.1227 liters).

Moles LiBr: (0.1227 liters)*(0.150 moles/liter) = 0.01841 moles LiBr

Since we only get 1 mole of PbBr₂ per 2 moles of LiBr, we'll get:

(0.01841 moles LiBr)*(1 mole PBr₂/2 moles LiBr) = 0.00920 moles PBr₂

Bob