Sodium carbonate reacts with silver nitrate according to the following balanced equation: Na2CO3 (s) + 2 AgNO3 (aq) → Ag2CO3 (s) + 2 NaNO3 (9) If 3.60 g of Na2CO3 is allowed to react with 5.14 g of AgNO3, what mass of Na2CO3 will remain at the end of the reaction? a) 2.01 g
b) 1.54 g
c) 0.423 g
d) 0.0150 g
e) 0g

a) 2.01 g

Explanation:

First we convert 0.0302 mol AgNO₃ to Na₂CO₃ moles, in order to calculate how many Na₂CO₃ moles reacted:

So the remaining Na₂CO₃ moles are:

Finally we convert Na₂CO₃ moles into grams, using its molar mass:

The closest answer is option a).

you did not give any options thus this is the best answer i can give to you with this little information you have given

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