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scZoUnD [109]
3 years ago
7

Calculate the number of moles of excess reactant that will be left-over when 50.0 g of KI react with 50.0 g of Br2: 2KI + Br2 2K

Br + I2

Chemistry
2 answers:
Ugo [173]3 years ago
4 0
Hope this helps you.

kvv77 [185]3 years ago
3 0

Answer : The moles of excess reactant, Br_2 is, 0.1625 mole.

Explanation : Given,

Mass of KI = 50 g

Mass of Br_2 = 50 g

Molar mass of KI = 166 g/mole

Molar mass of Br_2 = 160 g/mole

First we have to calculate the moles of KI and Br_2.

\text{Moles of }KI=\frac{\text{Mass of }KI}{\text{Molar mass of }KI}=\frac{50g}{166g/mole}=0.301moles

\text{Moles of }Br_2=\frac{\text{Mass of }Br_2}{\text{Molar mass of }Br_2}=\frac{50g}{160g/mole}=0.313moles

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

2KI+Br_2\rightarrow 2KBr+I_2

From the balanced reaction we conclude that

As, 2 moles of KI react with 1 mole of Br_2

So, 0.301 moles of KI react with \frac{0.301}{2}=0.1505 moles of Br_2

From this we conclude that, Br_2 is an excess reagent because the given moles are greater than the required moles and KI is a limiting reagent and it limits the formation of product.

The moles of excess reactant, Br_2 = Given moles - Required moles

The moles of excess reactant, Br_2 = 0.313 - 0.1505 = 0.1625 mole

Therefore, the moles of excess reactant, Br_2 is, 0.1625 mole.

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