In the laboratory, a quantity of I2 was reacted with excess H2 to give 0.650 moles...

In the laboratory, a quantity of I2 was reacted with excess H2 to give 0.650 moles of HI. It is also known that the percent yield of this reaction was 46.0%. How many moles of I2 reacted? (Hint: Write a balanced equation.)

Expert Solution

H2 + I2 ---------> 2HI

mass of HI = no of moles* molar mass

= 0.65*128 = 83.2g of HI

percentage yield = actual yield*100/theoretical yield

46 = 83.2*100/theoretical yield

Theoreticla yield = 83.2*100/46 = 180.87g

H2 + I2 ---------> 2HI

2 moles of HI formed from 1 mole of I2

2*128g of HI formed from 1 mole of I2

180.87g of HI formed from = 1mole*180.87/2*128 = 0.7065 moles of I2

queen_honey_blossom answered 10 months ago

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In the laboratory, a quantity of I2 was reacted with excess H2 to give 0.650 moles...

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