Question

10. 1.29 dag of sodium chloride is reacted with 25.0 dag of silver nitrate10. Balanced Chemical Equation: Reaction Type:At completion of reactions: Formula of Reactant A: Grams of Reactant A: Formula of Reactant B:Grams of Reactant B: Formula of Product C:Grams of Product C: Formula of Product D:Grams of Product D:

Answer 1

So,

Let's first convert some units:

`\begin{gathered} 1.29dag\cdot(10g)/(1dag)=12.9g \\ \\ 25.0\text{dag}\cdot\frac{10g}{1\text{dag}}=250g \end{gathered}`

So,

There's a reaction between 12.9g of sodium chloride and 250g of silver nitrate.

The reaction that occurs here, is the next one:

`NaCl+AgNO_3\to AgCl+NaNO_3`

Sodium chloride reacts with silver nitrate to form silver chloride and sodium nitrate.

This reaction is a displacement reaction.

Now, let's complete the information:

Formula of Reactant A: NaCl

Grams of Reactant A: 12.9g

Formula of Reactant B: AgNO3

Grams of Reactant B: 250g

Formula of Product C: AgCl

Grams of Product C: ?

Formula of Product D: NaNO3

Grams of Product D: ?

We need to find the amount of grams of the formed products.

So, let's first convert the grams of each reactant in moles. Remember that the number of moles (n) can be found if we divide the given mass of the given compound by its molar mass. So,

`\begin{gathered} n_(NaCl)=\frac{12.9gNaCl}{\frac{58.5gNaCl}{\text{mol}}}=0.22\text{molesNaCl} \\ _{} \\ n_(AgNO_3)=(250gAgNO_3)/((169.87gAgNO_3)/(mol))=1.47molesAgNO_3 \end{gathered}`

The limiting reactant in this reaction is NaCl. So, we're going to use the reaction's stoichiometry to find the amount of the products as follows:

`0.22mol\text{NaCl}\cdot\frac{1\text{molAgCl}}{1\text{molNaCl}}\cdot\frac{143.32\text{gAgCl}}{1\text{molAgCl}}=31.53\text{gAgCl}`

Thus, there are 31.53g of AgCl formed.

Finally, let's find the amount of grams of product D (NaNO3):

`0.22mol\text{NaCl}\cdot\frac{1\text{molNaNO}3}{1\text{molNaCl}}\cdot\frac{84.99\text{gNaNO}3}{1\text{molNaNO}3}=18.69\text{gNaNO}3`

Therefore,

There were formed 31.53g of AgCl (product C) and 18.69g of NaNO3 (product D)