Question

2 Pb(NO3)2 --> 2 PbO + 4 NO2 + O2  

 

 

How many grams of lead (II) oxide are produced to form 5 grams of oxygen? 

 

 

 

 

 

 

How many grams of lead (II) nitrate are needed to make 7 moles of nitrogen dioxide  

 

 

 

 

 

 

 

How many moles of oxygen can be made from 14 moles of lead (II) nitrate?  

Answer 1

The given thermal decomposition reaction can be shown as follows

$2Pb(NO_3)_2 \longrightarrow2PbO+4NO_2+O_2$

from above reaction we can say that 2 mole of lead nitrate gives 2 mole of lead oxide, 4 mole of nitrogen dioxide and 1 mole of oxygen

(a)

mole of oxygen produced = $\frac{5}{32}=0.15625$

mole of lead oxide produced = $2 \times 0.15625= 0.3125$

mass of lead oxide = $223.2\times 0.3125 = 69.75 g$

(b) 4 mole of nitrogen dioxide produced from 2 mole of lead oxide

7 mole of dinitrogen produced from = $\frac{2}{4}\times 7 = 3.5$ lead nitrate required

Mass of lead nitrate = 1159.2 g

(c) 2 mole of leadnitrate gives 1 mole of dioxygen

Now, 14 mole of lead nitrat gives 7 mole of dioxygen