
If x g weight of will be precipitated when a solution containing 4.77 g is added to a solution of 5.77 g of , then find the value of x.
A. 143.5
B. 142.5
C. 147.5
D. None of the above
Answer
488.4k+ views
Hint: Write the balanced chemical reaction for the production of using and . 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 . Finally, convert the moles product to the mass of the product.
Complete Step by step answer: We have given that will be precipitated when reacts with .
So first we will write a balanced chemical reaction.
Now, using the given masses of and we will calculate the moles of both the reactants.
Mass of = 4.77 g
Molar mass of = 58.44 g/mol
So,
Similarly, we can calculate the moles of as follows:
Mass of = 5.77 g
Molar mass of = 169.87 g/mol
So,
Form the balanced chemical equation we can say that the mole ratio of : : is 1:1:1
As we have 0.0816 mol of and 0.03396 mol of so we can say 0.03396 mol of will react with 0.03396 mol of and will give 0.03396 mol of
Here is the limiting reagent so we have determined the moles of product from moles of limiting reagent.
Now, we have moles of . Now we have to convert these moles of to mass using its molar mass as follows:
Molar mass of = 143.5 g/mol
Thus, we can say that 4.87 g of will be precipitated out when a solution containing 4.77 g is added to a solution of 5.77 g of .
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.
Complete Step by step answer: We have given that
So first we will write a balanced chemical reaction.
Now, using the given masses of
Mass of
Molar mass of
So,
Similarly, we can calculate the moles of
Mass of
Molar mass of
So,
Form the balanced chemical equation we can say that the mole ratio of
As we have 0.0816 mol of
Here
Now, we have moles of
Molar mass of
Thus, we can say that 4.87 g of
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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