Question

Which of the following laboratory procedures best illustrates the law of conservation of mass?

Answer 1

Answer: The correct statement is reacting 12 g of C with 32 g of $O_2$ to form 44 g of $CO_2$ and unused reactants.

Explanation:

Law of conservation of mass states that mass can neither be created nor be destroyed, but it can only be transformed from one form to another. The total mass in the reaction remains constant.

From the given options:

• 1. Reacting 12 g of C with 32 g of $O_2$ to form 44 g of $CO_2$ and unused reactants

The equation for this is given by:

$C+O_2\rightarrow CO_2$

Mass on the reactant side = (12 + 32)g = 44 g

Mass on the product side = 44 g

Hence, this equation follows Law of Conservation of mass.

• 2. Weighing 100 g of Cu powder and 100 g of Fe filings

As, there is no equation formed, so there will be no chemical reaction and will not follow, law of conservation of mass.

• 3. Heating 100 g of $CaCO_3$ to produce 56 g of CaO

The equation for this is given by:

$CaCO_3\rightarrow CaO+CO_2$

Mass on the reactant side = 100 g

Mass on the product side = (56 + 0)g = 56g

Hence, this equation does not follow Law of Conservation of mass.

• 4. Dissolving 10 g of NaCl in 100 g of water

As, salt is getting dissolved in water and hence, it is a physical change and will not follow law of conservation of mass.

Therefore, the correct statement is reacting 12 g of C with 32 g of $O_2$ to form 44 g of $CO_2$ and unused reactants.

Answer 2

The  laboratory  procedure that best  illustrate the law of conservation  is

heating 100 g of CaCo3  to produce  56 g of CaO  (answer C)

explanation

According to the law  of mass conservation ,  the mass of the  reactant  must   be equal  to the mass  of the product.

 According  to  option c Heating  100 g CaCO3  to produces  56 g CaO  (  40 +16=56)

The remaining mass  = 100-56  =  44  which  would the mass of CO2  [  12 + (16 x2)]= 44   since  CaCO3  decomposes to produce CaO  and  CO2

Therefore  the  mass  of reactant=   100g

                       mass  of product =  56 g +44 g =100

Therefore the laboratory    procedure for  decomposition  of CaCO3  illustrate the law of mass  conservation since the mass of reactant = mass of product.