How many grams of Ag2CO3 are required to reacr with 28.5 mL of 1.00 M NaOH solution?

1 answer:

8 0

The reaction between Ag2CO3 and NaOH is shown by the equation below
Ag2CO3 + NaOH = Ag2O + Na2CO3 +H2O
we can determine the number of mole of sodium hydroxide
by (2.85 ml × 1) ÷ 1000 ml , since according to molarity 1 mole is contained in 100ml.
we get 0.00285 moles of NaOH
Using the mole ratio we can get the moles of Ag2CO3
Mole ratio: Ag2NO3 : NaOH = 1:1
Therefore, the moles of Ag2CO3 will be 0.00285 moles
but 1 mole of silver carbonate is equivalent to 275.8 g
Thus the mass will be calculated by 0.00285 moles × 275.8g = 0.78603g
Mass of silver carbonate required will be 0.78603g

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