Question

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

Answer 1

Answer:

a) 2.01 g

Explanation:

• Na₂CO₃ (s) + 2AgNO₃ (aq) → Ag₂CO₃ (s) + 2NaNO₃

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

• 0.0302 mol AgNO₃ * $\frac{1molNa_2CO_3}{2molAgNO_3}$  = 0.0151 mol Na₂CO₃

So the remaining Na₂CO₃ moles are:

• 0.0340 - 0.0151 = 0.0189 moles Na₂CO₃

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

• 0.0189 moles Na₂CO₃ * 106 g/mol = 2.003 g Na₂CO₃

The closest answer is option a).