Question

The number of moles of CaCl2 needed to react with excess AgNO3 to produce 4.31 gram AgCl is

Answer 1

Answer:

0.015

Explanation:

Answer 2

Answer:

1.665g of $CaCl_{2}$ is required to produce 4.31g of $AgCl$.

Explanation:

$CaCl_{2} + 2AgNO_{3}$ → $2AgCl + Ca(NO_{3})_{2}$

One mole $CaCl_{2}$ reacts with two moles of $AgNO_{3}$ to produce two moles of $AgCl$ and one mole of $Ca(NO_{3})_{2}$.

We have to find out the number of moles of $CaCl_{2}$ needed to produce 4.31g of $AgCl$.

Find we convert the mass of $AgCl$ to moles by using the formula below-

$Moles=\frac{Mass}{Molar Mass}$

The molar mass of $AgCl$ = 143g/mol

The mass of $AgCl$= 4.31g

$Moles = \frac{4.31}{143}= 0.03mol$ of $AgCl$

We know that:

1 mole of $CaCl_{2}$ → 2 moles of$AgCl$

$x$ mole of $CaCl_{2}$ → 0.03 moles of $AgCl$

$2 * x = 1*0.03$

$x = \frac{0.03}{2}$

$x=0.015 mol$ of $CaCl_{2}$.

Convert the moles into mass using the same formula.

The moles of $CaCl_{2}$- 0.015mol

The molar mass of $CaCl_{2}$- 111g/mol

$0.015=\frac{mass}{111}$

$1.665g =mass$

Conclusion:

The mass required to produce 4.31g of$AgCl$ is 1.665g.