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Equal weight of Zinc metal and Iodine are mixed together and the Iodine is completely converted to zinc iodide What fraction of weight of the original Zinc remains unreacted?
Wiki User
∙ 11y ago
Balanced equation is:
Zn(s) + I2(s) --> ZnI2(s)
Pick any initial mass of Zn and I2, and convert those masses to moles:
100 g Zn / 63.4 g/mol = 1.58 moles
100 g I2 / 253.8 g/mol = 0.394 mol I2
Since these two react in a 1:1 ratio, you will have used 0.394 moles of Zn to react with the 0.394 moles of I2. That would leave 1.58 - 0.394 = 1.186 moles of Zn unreacted.
So, the fraction of the original zinc remaining would be 1.186 / 1.58 = .75
by sonu gupta
Wiki User
∙ 11y ago
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