The balanced Equation Is: Na2CO3(aq)+2AgNO3 (aq) --> 2NaNO3(aq)+Ag2CO3 (s)------

Question

The balanced Equation Is: Na2CO3(aq)+2AgNO3 (aq) --> 2NaNO3(aq)+Ag2CO3 (s)------------------------------------------------------------------------------------------------------------------------------------------ Question 1: 10 mL of AgNO3 in this contains 34 grams of AgNO3. Calculate the grams of solid product of Ag2CO3.-------------------------------------------------------------------------------------------------Question 2: 10 mL of Na2CO3 in this contains 21 grams of Na2CO3. Calculation that the grams of solid product of Ag2CO3.

Explanation / Answer

1) 34 gm AgNO3 = 34/169.87 = 0.20015 moles (where 169.87 is mol wt of AgNO3) from eq it is clear that 2 moles AgNO3 gave 1 mole Ag2CO3 hence 0.20015 moles AgNO3 gives (0.20015/2) = 0.100076 moles Ag2CO3 mass of Ag2CO3 = 0.100076 x275.7453 = 27.5956 gm = 27.6 grams (where 275.7453 is mo wt of Ag2CO3) 2) 21 grams Na2CO3 = 21/105.9784 = 0.19815 moles 1 mole Na2CO3 gave 1 mole Ag2CO3 0.19815 moles Na2CO3 gives 0.19815 moles Ag2CO3 mass of AG2CO3 = 0.19815 x 275.7453 = 54.64 grams

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