Question

9Na + 4ZnI2 = 8NaI + NaZn4

From this equation:

If 526,68 grams of ZnI2 is completely reacted, what is the mass of NaZn4 produced? (Show full working)

Answer 1

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.