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Bond [772]
3 years ago
13

The decomposition of 57.0 g of Fe2O3 results in Consider the following reaction. 2Fe2O3 ---> 4Fe + 3O2 deltaH degree rxn = +

824.2 kJ decomposition of 57.0 g of Fe2O3 results in the release of 294 kJ of heat. A. the absorption of 23500 kJ of heat. B. the absorption of 147 kJ of heat. C. the absorption of 294 kJ of heat. D. the release of 23500 kJ of heat. E. the release of 147 kJ of heat.
Chemistry
1 answer:
lozanna [386]3 years ago
3 0

Answer:

The correct answer is option E.

Explanation:

2Fe_2O_3\rightarrow 4Fe + 3O_2 \Delta H^o_{rxn} = 824.2 kJ

Mass of ferric oxide decomposed = 57.0 g

Moles of ferric oxide decomposed = \frac{57.0 g}{160 g/mol}=0.35625 mol

According to reaction, 2 moles ferric oxide on combustion gives 824.2 kJ of heat.

Then heat given by on decomposition of 0.35625 mol of ferric oxide will be:

\frac{824.2 kJ}{2}\times 0.35625 =146.81 kJ\approx 147 kJ

So, on decomposition of 57.0 grams of ferric oxide 147 kilo Joules of heat was released.

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Q: A
according to this formula, we can get the mole fraction of water (n):
P(solu) = n Pv(water)
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22.8 = n * 23.8
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moles of water = 500 g(mass weight) / 18 (molar weight)= 27.7 mol
n = moles of water / ( moles of water + moles of glucose)
0.958   = 27.7 / ( 27.7+ moles of glucose)
0.958 moles of glucose + 26.5 = 27.7
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here we also need to get n (mole fraction of water )by using this formula:
Pv(solu) = n Pv(water)
when we have Pv(solu)=132 & Pv(water)=150 so, by substition:
132= n * 150
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- total moles in solution = moles of water / moles fraction of water 
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                                       = 5.34 - 4.7 = 0.64 moles solute
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moles of urea (NH2)2 CO = mass weight / molar mass
                                           = 4.49 g / 60 g /mol
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moles fraction of methanol = moles of methanol / (moles of methanol + moles of urea )
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Pv(solu) = n P°v
Pv(solu) = 0.95 * 89 mm Hg 
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