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 :

The mass of excess mass of , are, 1.908 g, 0 g, 12.144 g and 3.74 g respectively.

Explanation : Given,

Mass of = 4.25 g

Mass of = 7.50 g

Molar mass of = 106 g/mole

Molar mass of = 170 g/mole

Molar mass of = 276 g/mole

Molar mass of = 85 g/mole

First we have to calculate the moles of and .

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

From the balanced reaction we conclude that

As, 2 moles of react with 1 mole of

So, 0.044 moles of react with moles of

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

The excess mole of = 0.040 - 0.022 = 0.018 mole

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

Now we have to calculate the moles of .

As, 1 moles of react to give 1 moles of

So, 0.044 moles of react to give 0.044 moles of

Now we have to calculate the mass of .

Now we have to calculate the moles of .

As, 2 moles of react to give 2 moles of

So, 0.044 moles of react to give 0.044 moles of

Now we have to calculate the mass of .