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Question: How much mass of silver nitrate will react with 5.85 gm of Sodium Chloride to produce 14.35 gm of Si...

How much mass of silver nitrate will react with 5.85 gm of Sodium Chloride to produce 14.35 gm of Silver Chloride and 8.5 gm of Sodium nitrate if the law of conservation of mass is followed ?
(A) 22.85 gm
(B) 108 gm
(C) 17.0 gm
(D) 28.70 gm

Explanation

Solution

Hint: As a law of conservation of mass, total mass of reactants is always equal to total mass of products. To find the total mass of the reactants, we can do a sum of the weights of all the reactants expressed in the same units and the same procedure could be followed for finding total mass of products respectively.

Complete answer :
Below is the reaction and the molecular masses of the particular molecules.
AgNO3?+NaCl5.85gmAgCl14.35gm+NaNO38.5gm\mathop {AgN{O_3}}\limits_? + \mathop {NaCl}\limits_{5.85gm} \to \mathop {AgCl}\limits_{14.35gm} + \mathop {NaN{O_3}}\limits_{8.5gm}
Now, According to Law of conservation of molecular mass, no mass is lost during the course of the reaction and the total mass of the reactants is equal to the total mass of the products.
So, if we make the equation of the law of conservation of mass it will be like
Total mass of Reactants = Total mass of Products…………………..(1)
Putting the values according to reaction in the formula, we will get
Mass of AgNO3AgN{O_3}+ Mass of NaClNaCl=Mass of AgClAgCl+ Mass of NaNO3NaN{O_3}
Mass of AgNO3AgN{O_3}+ 5.85 gm = 14.35 gm + 8.5 gm…………………(2)
Simplifying the equation (2) we get
Mass of AgNO3AgN{O_3}=14.35 gm +8.5 gm – 5.85 gm
Mass ofAgNO3AgN{O_3}=17.0 gm

So, the correct choice of option will be (C) 17.0 gm.

Additional Information:
- In some cases the information of mass could be given in moles and if so then we need to convert them into mass by specific formula of moles.
number of moles n = weight of compoundmolecular weight of the compound{\text{number of moles n = }}\dfrac{{{\text{weight of compound}}}}{{{\text{molecular weight of the compound}}}}
e.g. If we need to find the number of moles of 40gm of NaOH, then we can find it as below.
number of moles n = 40gm40gmmol1{\text{number of moles n = }}\dfrac{{40gm}}{{40gmmo{l^{ - 1}}}}
Hence we can say that the number of moles in that case will be n=1.

Note: When a reaction involves two or more molecules of a compound, then accordingly we also need to consider them in the calculation of total mass. These types of questions just involve simple addition and subtraction of masses of reactants and products, so do not make any mathematical error.