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A theoretical yield is the amount of substance you calculate mathematically, as opposed to the actual yield, which is the amount of substance you find using a balance.

To find the theoretical yield, you need a balanced equation. In this case,

Zn + I2 -> ZnI2

So, that was easy enough. The theoretical yield is limited by the reagent present in the smaller quantity by moles, not by mass. You must take the mass you have of zinc multiplied by zinc's molar mass to find the number of moles of zinc. You must take the mass you have of I2 and multiply by the molar mass of I2.

The limiting reagent is the substance of fewer moles. Since these two reagents react in a 1:1 ratio, you just need to compare and see which you have less of. This will give you the number of moles you can make of ZnI2, theoretically -- the theoretical yield.

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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


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Related Questions

What is the balanced equation for zinc plus lead nitrate yield zinc nitrate plus lead?

Equation is Zn + Pb(NO3)2 --> Zn(NO3)2 + Pb


Zn plus MgCl2 yields what?

The reaction between zinc (Zn) and magnesium chloride (MgCl2) would yield zinc chloride (ZnCl2) and magnesium (Mg).


Write a chemical equation for zinc oxide plus carbon to yield zinc vapor and CO?

ZnO + C --> Zn + CO


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?

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


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What is the balanced equation for KOH plus Zn?

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