Question

Solutions of sodium carbonate and silver nitrate react to form solid silver carbonate and a solution of sodium nitrate. A solution containing 4.25 g of sodium carbonate is mixed with one containing 7.50 g of silver nitrate. How many grams of each of the following compounds are present after the reaction is complete?

i) sodium carbonate
ii) silver nitrate
iii) silver carbonate
iv) sodium nitrate

Answer 1

To determine the grams of each compound present after the reaction, we need to use stoichiometry and the balanced equation of the reaction.

Step-by-step explanation:

To determine the grams of each compound present after the reaction, we need to use stoichiometry and the balanced equation of the reaction. The balanced equation is:

AgNO3(aq) + Na2CO3(aq) → Ag2CO3(s) + 2NaNO3 (aq)

1. To find the grams of sodium carbonate remaining, we need to calculate the moles of sodium carbonate reacted and then convert it to grams using its molar mass.
2. To find the grams of silver nitrate remaining, we need to calculate the moles of silver nitrate reacted and then convert it to grams using its molar mass.
3. To find the grams of silver carbonate formed, we need to calculate the moles of silver carbonate using the stoichiometric ratio from the balanced equation, and then convert it to grams using its molar mass.
4. To find the grams of sodium nitrate formed, we need to calculate the moles of sodium nitrate using the stoichiometric ratio from the balanced equation, and then convert it to grams using its molar mass.

Answer 2

The mass of excess mass of
`Na_2CO_3`,
`AgNO_3,Ag_2CO_3\text{ and }NaNO_3` are, 1.908 g, 0 g, 12.144 g and 3.74 g respectively.

Explanation : Given,

Mass of
`Na_2CO_3` = 4.25 g

Mass of
`AgNO_3` = 7.50 g

Molar mass of
`Na_2CO_3` = 106 g/mole

Molar mass of
`AgNO_3` = 170 g/mole

Molar mass of
`Ag_2CO_3` = 276 g/mole

Molar mass of
`NaNO_3` = 85 g/mole

First we have to calculate the moles of
`Na_2CO_3` and
`AgNO_3`.

`\text{Moles of }Na_2CO_3=\frac{\text{Mass of }Na_2CO_3}{\text{Molar mass of }Na_2CO_3}=(4.25g)/(106g/mole)=0.040moles`

`\text{Moles of }AgNO_3=\frac{\text{Mass of }AgNO_3}{\text{Molar mass of }AgNO_3}=(7.50g)/(170g/mole)=0.044moles`

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

`Na_2CO_3+2AgNO_3\rightarrow Ag_2CO_3+2NaNO_3`

From the balanced reaction we conclude that

As, 2 moles of
`AgNO_3` react with 1 mole of
`Na_2CO_3`

So, 0.044 moles of
`AgNO_3` react with
`(0.044)/(2)=0.022` moles of
`Na_2CO_3`

From this we conclude that,
`Na_2CO_3` is an excess reagent because the given moles are greater than the required moles and
`AgNO_3` is a limiting reagent and it limits the formation of product.

The excess mole of
`Na_2CO_3` = 0.040 - 0.022 = 0.018 mole

Now we have to calculate the mass of excess mole of
`Na_2CO_3`.

`\text{Mass of }Na_2CO_3=\text{Moles of }Na_2CO_3* \text{Molar mass of }Na_2CO_3=(0.018mole)* (106g/mole)=1.908g`

Now we have to calculate the moles of
`Ag_2CO_3`.

As, 1 moles of
`AgNO_3` react to give 1 moles of
`Ag_2CO_3`

So, 0.044 moles of
`AgNO_3` react to give 0.044 moles of
`Ag_2CO_3`

Now we have to calculate the mass of
`AgCO_3`.

`\text{Mass of }Ag_2CO_3=\text{Moles of }Ag_2CO_3* \text{Molar mass of }Ag_2CO_3=(0.044mole)* (276g/mole)=12.144g`

Now we have to calculate the moles of
`NaNO_3`.

As, 2 moles of
`AgNO_3` react to give 2 moles of
`NaNO_3`

So, 0.044 moles of
`AgNO_3` react to give 0.044 moles of
`NaNO_3`

Now we have to calculate the mass of
`NaNO_3`.

`\text{Mass of }NaNO_3=\text{Moles of }NaNO_3* \text{Molar mass of }NaNO_3=(0.044mole)* (85g/mole)=3.74g`