Question

Cuantos gramos de cloruro de plata se puede preparar a partir de 620 gramos de nitrato de plata

Answer 1

AgNO3 + NaCl → NaNO3 + AgCl

Mm AgNO3 = 169,87 g/mol

AgCl = 143,32 g/mol

g AgCl = 620 g AgNO3 x 1 mol AgNO3 x 1 mol Ag Cl x 143.32 g AgCl

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169.87 g AgNO3 1 mol AgNO3 1 mol Ag Cl

g AgCl = 523.10