Check of mass calculation?

Question

2AgNO₃(aq) KrCrO₄(aq) --> Ag₂CrO₄(s) 2KNO₃(aq)

0.778g of precipitate formed.

Mass of silver nitrate that reacted?

n(Ag₂CrO₄) = 0.00235

M(2AgNO₃) = 2(107.87) 2(14) 6(16) = 339.74

2AgNO₃: Ag₂CrO₄ ∴ 2 : 1 ratio

2(339.74) x 0.00235 = 1.597g of AgNO₃ reacted.

Thanks for your time.

Answer 1

2AgNO₃(aq) + KrCrO₄(aq) --> Ag₂CrO₄(s) + 2KNO₃(aq)

Ag: 107.9

Cr:  52

O:  16

N:  14

AgNO3: 169.9

Ag2CrO4: 331.8

0.778g of precipitate formed. Of this 0.778*215.8/331.8 = 0.506 g was silver. The amount of AgNO3 containing this quantity of silver is given by 0.506*169.9/107.9 = 0.7967 g

This is half your answer because you counted the "2" twice, that is, 339.74 already has taken the 2AgNO3 into account and you did that again with the 2:1 ratio

Answer 2

Moles Ag2CrO4 = 0.00235 ( ok)

moles AgNO3 = 2 x 0.00235= 0.00470

Molar mass AgNO3 = 169.88 g/mol

( you have not multiply by 2)

mass AgNO3 = 169.88 g/mol x 0.00470 = 0.798 g