Question

Review Part B Solutions of sodium carbonate and silver nitrate react to form solid silver carbonate and a solution of sodium nitrate. A solution containing 3.30 g of sodium carbonate is mixed with one containing 4.71 of silver nitrate How many grams of silver nitrate are present after the reaction is complete? V AE R O ? Submit Request Answer

Answer 1

Answer:

0 g

Explanation:

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Given: For sodium carbonate

Given mass = 3.30 g

Molar mass of sodium carbonate= 105.9888 g/mol

Moles of sodium carbonate = 3.30 g / 105.9888 g/mol = 0.03114 moles

Given: For silver nitrate

Given mass = 4.71 g

Molar mass of silver nitrate = 169.87 g/mol

Moles of silver nitrate = 4.71 g / 169.87 g/mol = 0.02773 moles

According to the given reaction:

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

1 mole of sodium carbonate react with 2 moles of silver nitrate

0.03114 moles  of sodium carbonate react with 2*0.03114 moles of silver nitrate

Moles of silver nitrate = 0.06228 moles

Available moles = 0.02773 moles

Limiting reagent is the one which is present in small molar amount. It got exhausted at the end of the reaction and the formation of the products is governed by it.

Thus, 0.02773 < 0.06228 moles, Silver nitrate is the limiting reagent and will not left at the end  of the reaction.