Determine the mass in grams of KBr formed from 10.13 grams of bromine gas.
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Answer:
Explanation:
In 100gKBr,K=32g. Then Br=67.1g
(at.wt. of K and Br are 39 and 80 respectively)
2K+Br2→2KBr
Mole ofKMole ofBr2=32.9/3967.1/80=1.00571
∵Moles ofKMoles ofBr2=1.0057×21=2.01141
Given moles2K3.6039=0.092+6.40160=0.04Br2→2KBr
The reaction ratio for moles of K and Br2=2.0114
Thus, Br2 will be completely used leaving K.
∵ Mole of Br2 reacting =0.04
∴ Moles of K reacting =0.04×2.0114
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