Identify the limiting and excess reactants when 1.00 g of zinc reacts with 150 ml of 0.250m pb(no3)2. how many grams of lead are formed in this single replacement reaction

The reaction equation is:
Zn + Pb(NO₃)₂ → Pb + Zn(NO₃)₂

The moles of lead are calculated using:
moles = concentration x volume
moles = 0.25 x 0.15
moles = 0.0375

The moles of zinc are: 1 / 65 = 0.015

We see from the equation that equimolar quantities of zinc and lead should be present. Therefore, lead is in excess and zinc is the limiting reactant.

The answers is bad germs in their

(no3)⁻n has oxidation number x,o has oxidation number -2.(x*1 -2*3) charge of the nitrate ion, and we know that charge of the nitrate ion =-1.so,(x*1 -2*3) =-1x-6=-1x=+5answer a. +5.