# What weight of AgCl will be precipitated when a solution containing 4.77g of NaCl is added to a solution of 5.77g of AgNO_3 ?

## I really don't get this question! I know how to solve a basic stoichiometry problem but this one is tough for me, so please help me with this so I can solve this kind of problem.

May 26, 2017

$4.87$ gram $A g C l$ will be precipitated.

#### Explanation:

The first thing to do is to write down the reaction that is described in the question. Since the reactant and one of the product are given, we can write the balanced reaction:

$N a C l + A g N {O}_{3} \to A g C l + N a N {O}_{3}$

In this type of questions, it is nice to write down a kind of 'scheme' in which we place all the information provided so we can have a clear look what to do. But first, it is important to calculate the masses into moles, before we on accident calculate with the wrong numbers. To do this, we need to molar masses of the compounds. This can be calculated by adding up the masses of the atoms in the compounds. We obtain:

$A g C l$=$1 \times A g + 1 \times C l = 107.9 u + 35.45 u = 143.35 u$
$N a C l$=$1 \times N a + 1 \times C l = 22.99 u + 34.45 u = 57.44 u$
$A g N {O}_{3}$=$1 \times A g + 1 \times N + 3 \times O = 107.9 u + 14.01 u + 3 \times 16.00 u = 169.91 u$

Now we calculate a number of moles with the formula below.

$m o l = \frac{m a s s}{m o l a r m a s s}$

I only display the calculation for $N a C l$ here.
$m o l = \frac{4.77 \left(\textcolor{red}{\cancel{\textcolor{b l u e}{g r a m}}}\right)}{57.44 \left(\frac{\textcolor{red}{\cancel{\textcolor{b l u e}{g r a m}}}}{\textcolor{b l u e}{m o l}}\right)} = 0.083 \textcolor{w h i t e}{a} \textcolor{b l u e}{m o l}$ $N a C l$
$A g N {O}_{3} = 0.034 \textcolor{w h i t e}{a} \textcolor{b l u e}{m o l}$

Now we have the moles, we write the molar ratio below the reaction and the moles at the correct places. We obtain something like this:

$\textcolor{w h i t e}{a a}$$N a C l + A g N {O}_{3} \to A g C l + N a N {O}_{3}$$\textcolor{w h i t e}{a a a} \textcolor{b l u e}{\text{reaction}}$
$\textcolor{w h i t e}{a a a a} 1 \textcolor{w h i t e}{a a b} : \textcolor{w h i t e}{a a a} 1 \textcolor{w h i t e}{a a a b} : \textcolor{w h i t e}{a a a} 1 \textcolor{w h i t e}{a a} : \textcolor{w h i t e}{a a a b} 1 \textcolor{w h i t e}{a a . a a} \textcolor{b l u e}{\text{molar ratio}}$
color(white)(aa)0.083 color(white)(aaa)0.034color(white)(aaaaaaa)?color(white)(aaaaaaaaaaaaa)color(blue)("moles")

We don't need to do anything with the $N a N {O}_{3}$ in this question.

We calculate the question mark ? by applying the law of conservation of mass. This can be done by:
mol $A g C l$ =(mol color(red)(AgNO_3)xxcolor(blue)(cancel(color(black)(molar ratio))) color(red)(AgCl))/(color(blue)(cancel(color(black)(molar ratio)) color(red)(AgNO_3))
mol $A g C l$ =$\frac{0.034 \times 1}{1} = 0.034$

Now we calculate the mass of $A g C l$ that is created. The molar mass was $143.35 u$

mass $A g C l$=$m o l \times m o l a r m a s s$
mass $A g C l$ =$0.034 \textcolor{red}{\cancel{\textcolor{b l u e}{m o l}}} \times 143.35 \frac{\textcolor{b l u e}{g r a m}}{\textcolor{red}{\cancel{\textcolor{b l u e}{m o l}}}} = 4.87 \textcolor{w h i t e}{a} \textcolor{b l u e}{g r a m}$

For this particular question, it seems useless to calculate the molar mass of the other reactant, but for future questions, they may ask something more about the reaction and then you already have made a nice scheme.