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Question

# Like most alkali metals, potassium reacts with water to form basic potassium hydroxide and hydrogen gas according to the reaction, 2K+2H2O→2KOH+H2 What amount of hydrogen gas will be formed when 117 g of potassium is added to 72 g of water?

A
1.61 g
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B
3.00 g
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C
4.73 g
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D
6.44 g
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Solution

## The correct option is B 3.00 gMoles of K = given massmolar mass=11739=3 moles Moles of H2O=given massmolar mass=7218=4 moles According to the reaction, 2 moles of K require 2 moles of H2O. Therefore, 3 moles of K will require 3 moles of H2O From the above calculations we can see that 1 mole of H2O is in excess. Therefore, potassium is the limiting reagent. Now, 2 moles of K produce 1 mole of H2 3 moles of K will produce 12×3=1.5 moles of H2 So, mass of H2 produced =(1.5×2) g = 3.0 g

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