How many grams ofsodium carbonate are present after the reaction iscomplete? How

Question

How many grams ofsodium carbonate are present after the reaction iscomplete?
How many grams ofsilver nitrate are present after the reaction iscomplete?
How many grams ofsilver carbonate are present after the reaction iscomplete?
How many grams ofsodium nitrate are present after the reaction iscomplete?
If you could help mework through these problems, I would be thrilled of sodium carbonate is mixed withone containing 5.00 of silver nitrate. How many grams ofsodium carbonate are present after the reaction iscomplete? How many grams ofsilver nitrate are present after the reaction iscomplete? How many grams ofsilver carbonate are present after the reaction iscomplete? How many grams ofsodium nitrate are present after the reaction iscomplete? If you could help mework through these problems, I would be thrilled Here is the problem: Solutions of sodium carbonate and silvernitrate react to form solid silver carbonate and a solution ofsodium nitrate. A solution containing 3.50

Explanation / Answer

The balanced equation of the given reaction is - Na2CO3 (aq) + 2AgNO3 (aq) ------> Ag2CO3 (s) + 2 NaNO3(aq) weight of sodium carbonate in the solution = 3.50g Therefore number of moles of sodium carbonate = 3.50 g /105.98g/mol                                                                           =0.033 mol Similarly number of moles of the silver nitrate = 5.0g/169.87 g/mol                                                                     = 0.0294 mol According to balanced equation we have 1 mol of  Na2CO3 for 2 moles ofAgNO3 . Hence from the above results we conclude thatAgNO3 is a limiting reactant. a) number of moles of Na2CO3reacted with AgNO3              = 0.0294 mol * (1 mol Na2CO3 /2 molAgNO3 )*(105.98 g/1 molNa2CO3)              = 1.5579 g Thus the mass of Na2CO3remained =3.50 g - 1.5579 g                                                       = 1.942 g b) Silver nitrate in the reactionis limiting, Hence there is no silver nitrate is presentafter completion of the reaction.                 = 0.0294 mol * (1 mol Na2CO3 /2 molAgNO3 )*(105.98 g/1 molNa2CO3)              = 1.5579 g Thus the mass of Na2CO3remained =3.50 g - 1.5579 g                                                       = 1.942 g b) Silver nitrate in the reactionis limiting, Hence there is no silver nitrate is presentafter completion of the reaction.   

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