How many grams of sodium carbonate are present after the reaction is complete?

The balanced equation would be 2AgNO3 + Na2CO3 --> Ag2CO3 + 2NaNO3

So 2 moles of AgNO3 would be needed to completely react with 1 mole of Na2CO3

The atomic weights of each of the above compounds are as follows....

AgNO3..................170

Na2CO3................106

Ag2CO3................276

NaNO3...................85

3.3/106 = 0.031132 moles of Na2CO3 are present and according to the equation, will require 2x0.031132= 0.062264 moles of AgNO3 (10.58 grams) for a complete reaction.

There are 4.71 grams of AgNO3 present, or 4.71/10.58 = 44.5% or the needed amount. This implies that 44.5% of the complete reaction will take place.

There is not enough AgNO3 for a complete reaction, so it looks like there will be no AgNO3 left when the reaction is complete.

There is more than enough Na2CO3 for a complete reaction, so it looks like there will be (1 - 44.5% = 55.5%) of Na2CO3 remaining when the reaction is complete:

55.5% x 3.3 grams = 1.8315 grams Na2CO3 remaining

44.5% x 3.3 grams = 1.4685 grams Na2CO3 used

convert to moles --> 1.4685 / 106 moles = 0.013853774 moles of Na2CO3 used

referring back to the right-hand side of the equation, the resulting products are:

...... --> Ag2CO3 + 2NaNO3

At this point, simply multiply the molecular weights by the moles used

0.013853774(276) Ag2CO3 = 3.8236 grams Ag2CO3

0.013853774(2x85) NaNO3 = 2.3551 grams NaNO3

Verify by comparing the weights of the reacted materials with the weights of the products; they should be equal.

4.71 grams AgNO3 + 1.4685 grams Na2CO3 --> 3.8236 grams Ag2CO3 + 2.3551 grams NaNO3

4.71 + 1.4685 =? 3.8236 + 2.3551

6.1785 =? 6.1788

Verified! (results do not match exactly due to rounding of decimals, but are close enough).

A bit long-winded, but hopefully some of it will make sense.

ANSWERS

Grams of sodium carbonate present after the reaction is complete:.....1.8315

Grams of silver nitrate present after the reaction is complete:.............NONE

Grams of sodium nitrate present after the reaction is complete:..........2.3551

Grams of silver carbonate present after the reaction is complete:........3.8236

tricky matter. query using the search engines. that could actually help!