What mass of silver chloride can be produced from 1.09 L of a 0.118 M solution of silver nitrate?

Question

Cameron Andrews · Accepted Answer

Under #20*g# of the halide may be precipitated.....

Our goal is to carry out a precipitation reaction. #Ag^(+) + Cl^(-)rarrAgCl(s)darr# And thus moles of #Ag^(+)-=Ag# We have an molar amount of... #1.09*Lxx0.118*mol*L^-1=0.129*mol# with respect to #Ag^+#. And thus an equimolar quantity of #AgCl# may be precipitated, which is as soluble as a brick (but much more difficult to handle!). Thus, the largest possible mass of "silver chloride" is... #143.32*g·mol^-1xx0.129*mol-=??g#

Jeremiah Adams · Answer

undefined To calculate the mass of silver chloride produced, we need to use the stoichiometry of the balanced chemical equation for the reaction between silver nitrate ($AgNO_3$) and sodium chloride ($NaCl$) to form silver chloride ($AgCl$).

First, we write the balanced chemical equation:
$$AgNO_3 + NaCl \rightarrow AgCl + NaNO_3$$

From the equation, we see that 1 mole of silver nitrate reacts with 1 mole of sodium chloride to produce 1 mole of silver chloride.

Given:
Volume of silver nitrate solution ($V$) = 1.09 L
Molarity of silver nitrate solution ($M$) = 0.118 M

Using the formula:
$$Molarity = \frac{moles}{volume}$$

We can find the moles of silver nitrate:
$$moles_{AgNO_3} = M \times V$$

Now, since the ratio of moles of silver nitrate to silver chloride is 1:1, the moles of silver chloride produced will also be $moles_{AgNO_3}$.

Now, we can use the molar mass of silver chloride ($AgCl$) to find the mass:
$$Mass_{AgCl} = moles_{AgCl} \times molar\ mass_{AgCl}$$

Given the molar mass of $AgCl$ is 143.32 g/mol.

Substitute the values and solve for the mass of silver chloride.