Solutions of sodium carbonate and silver nitrate react to form solid silver carbonate and a solution of sodium nitrate. A solution containing 3.40 g of sodium carbonate is mixed with one containing 4.86g of silver nitrate. Required:
How many grams of sodium carbonate, silver nitrate, silver carbonate, and sodium nitrate are present after the reaction is complete?

1.89 of Sodium Carbonate

3.94 g of Silver Carbonate

2.43 g of Sodium Nitrate

Zero grams of Silver Nitrate

Explanation:

We have to start with the reaction:

Now, we can balance the reaction:

Now, we have to calculate the limiting reagent and we have to follow a few steps:

1) Convert to moles (using the molar mass of each compound)

2) Divide by the coefficient of each reactive (given by the balanced reaction)

Convert to moles

Divide by the coefficient

The smallest value is for , therefore the 4.86 g of .

Now we can calculate the amount of compounds produced is we follow a few steps:

1) Use the molar ratio

2) Convert to moles (using the molar mass of each compound)

Amount of Silver Carbonate

Amount of Sodium Nitrate

Amount of Sodium Carbonate (Excess reactive)

Amount of Silver Nitrate

All the silver nitrate would be consumed in the reaction

I hope it helps!

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