Reaction: 9Na + 4ZnI2 = 8NaI + NaZn4
Mr (ZnI2) = 319 g/mol
Mr (NaZn4) = 283 g/mol
Let's make a proportion:
526,8/4•319 = m(NaZn4)/283
Annotation: coefficient 4 is from the reaction.
Hence, m(NaZn4) = 526,8•283/1276 = 116,837 g.
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