Answer in General Chemistry for thomas #86045

Question #86045

How many grams of Ag2CO3 will precipitate when excess Na2CO3 solution is added to 71.0 mL of 0.668 M AgNO3 solution?

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

______ g.

Expert's answer

2AgNO₃(aq) + Na₂CO₃(aq) $\to$ Ag₂CO₃(s)+2NaNO₃(aq)

n(AgNO₃)= $\dfrac{71mL*0.668 moles}{1000mL}=0.047428 moles$

n(Ag₂CO₃) = $\dfrac{0.047428moles}{2}= 0.023714moles$

m(Ag₂CO₃)= $0.023714moles*275.745 g/mol = 6.539g$

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Comments

Assignment Expert

21.10.20, 20:50

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danni

08.10.20, 19:54

Why do you divide 0.047428moles by 2 on the second step?