13. Exercises II

  

 

\(C\) + \(O_2\) → \(CO_2\)

Carbon 1 mole 12 g + Oxygen 1 mole 32 g → Carbon dioxide 42 gCarbon dioxide 44 gCarbon dioxide 48 gCarbon dioxide 43 g

 

It shows that 12 g of carbon burns in 32 g of oxygen form 44 g of carbon dioxide. Hence, 3 g 8 g 11 g 12 g of carbon reacts with 8 g of oxygen to form 11 g of carbon dioxide. It is given that 3.0 g of carbon is burnt with 8 g of oxygen to produce 11.0 g of \(CO_2\).

 

Therefore, 11.0 g of carbon dioxide will be formed when 3.0 g of \(C\) is burnt in 50 g of oxygen, consuming 8 g of oxygen, leaving behind 50 - 8 = 12 g 42 g44 g11 g of \(O_2\). The answer governs the law of gaseous volumeslaw of conservation of masslaw of multiple proportionlaw of constant proportion.