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nevsk [136]
3 years ago
13

100 g of an inorganic compound X.5H2O containing a volatile impurity was kept in an oven at 150 degree celcius for 60 minutes. t

he weight of residue after heating is 8g. the percentage of impurity in X was?
Chemistry
1 answer:
Hoochie [10]3 years ago
6 0
              150°C
X*5H₂O    →    X + 5H₂O↑

m₀=100 g
m(X)=8 g
M(H₂O)=18.0 g/mol
n(H₂O)=5 mol

the mass of the water 
m(H₂O)=M(H₂O)n(H₂O)
m(H₂O)=18.0*5=90 g

the mass of the residue after removal of water
m'(X)=100-90=10 g

the percentage of impurity in X*5H₂O
w(imp)=[m'(X)-m(X)]/m₀

w(imp)=[10-8]/100=0.02 (2%)
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1 mol  CH₃CH₂OCH₂CH₃ ______  74.12 g g

y                           ______   35.9 g

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100% yield _____ 0.5425 mol CH₃CH₂OCH₂CH₃

w                _____  0.48 mol CH₃CH₂OCH₂CH₃

w = 88.48%

b) Only 0.96 mol of ethanol reacted to form diethyl ether. This means that 0.125 mol of ethanol did not react. 45% of 0.125 mol reacted to form ethylene. Therefore, 0.05625 mol of ethanol reacted by the side reaction (reaction 2). Since 1 mol of ethanol leads to 1 mol of ethylene, 0.05625 mol of ethanol produces 0.05625 mol of ethylene.

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