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If x g weight of AgCl will be precipitated when a solution containing 4.77 g NaCl is added to a solution of 5.77 g of AgNO3, then find the value of x.
A. 143.5
B. 142.5
C. 147.5
D. None of the above

Answer
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Hint: Write the balanced chemical reaction for the production of AgCl using NaCl and AgNO3. From the given masses of the reactants calculate the moles of both the reactants. Determine the limiting reagent. Using the moles of the limiting reagent calculate the moles of the product AgCl. Finally, convert the moles product to the mass of the product.

Complete Step by step answer: We have given that AgCl will be precipitated when NaCl reacts with AgNO3 .
So first we will write a balanced chemical reaction.
NaCl + AgNO3 AgCl + NaNO3
Now, using the given masses of NaCl and AgNO3 we will calculate the moles of both the reactants.
moles = massmolar mass
Mass of NaCl = 4.77 g
Molar mass of NaCl = 58.44 g/mol
So, moles of NaCl = 4.77 g58.44 g/mol=0.0816 mole
Similarly, we can calculate the moles of AgNO3 as follows:
Mass of AgNO3 = 5.77 g
Molar mass of AgNO3 = 169.87 g/mol
So, moles of AgNO3 = 5.77 g169.87 g/mol=0.03396 mole
Form the balanced chemical equation we can say that the mole ratio of NaCl:AgNO3:AgCl is 1:1:1
As we have 0.0816 mol of NaCl and 0.03396 mol of AgNO3 so we can say 0.03396 mol of AgNO3 will react with 0.03396 mol of NaCl and will give 0.03396 mol of AgCl
Here AgNO3 is the limiting reagent so we have determined the moles of product from moles of limiting reagent.
Now, we have moles of AgCl. Now we have to convert these moles of AgCl to mass using its molar mass as follows:
Molar mass of AgCl = 143.5 g/mol
Mass of AgCl=moles \times molar mass
Mass of AgCl=0.03396 g ×143.5 g/mol
Mass of AgCl=4.87 g
Thus, we can say that 4.87 g of AgClwill be precipitated out when a solution containing 4.77 g NaCl is added to a solution of 5.77 g ofAgNO3.

Thus the correct option is (D) None of the above.

Note: Limiting reagent is the reagent which completely reacts in the reaction. So, the moles of limiting reagent and stoichiometric ratio of limiting reagent with product determine the amount of product. In mole concept people generally make mistakes in finding the exact solution due to unbalanced chemical equations.
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