Pb(no3)2 + 2kl > 2kno3 + pbi2
how many grams of lead (ll) nitrate is needed to completely react 25g or potassium iodide?

Mass = 24.84 g    

Explanation:

Given data:

Mass of potassium iodide = 25 g

Mass of lead nitrate required = ?

Solution:

Chemical equation:

Pb(NO₃)₂ + 2KI  →   2KNO₃ + PbI₂

Number of moles of potassium iodide:

Number of moles = mass/ molar mass

Number of moles = 25 g/ 166 g/mol

Number of moles = 0.15 mol

Now we will compare the moles of KI and Pb(NO₃)₂ .

                   KI           :          Pb(NO₃)₂

                    2           :               1

                   0.15       :             1/2×0.15 = 0.075 mol

Mass of lead nitrate:

Mass = number of moles × molar mass

Mass = 0.075 mol ×  331.2 g/mol

Mass = 24.84 g                

2h2o2--> 2h2o + o2

this reaction is of the spontaneous decomposition of hydrogen peroxide down into water and oxygen.

add 2 molecules of hydrogen peroxide and 2 molecules of water. since oxygen is naturally diatomic, the total number of atoms of each element is now the same on both sides of the equation so it is balanced.

this may not be entirely correct but let me know.