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

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