Thermogravimetric analysis and finding unknown value x
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Thermogravimetric analysis and finding unknown value x

[From: ] [author: ] [Date: 12-06-07] [Hit: ]
leaving 32.825g of solid residue.Molar mass of YBa2Cu3O(7−x) is (666.194 - 15.Mass equivalent of (0.5Y2O3 + 2BaO + 3Cu) is 610.......
YBa2Cu3O(7−x)(s) + ([ 7/2]−x)H2(g) →
[1/2]Y2O3(s) + 2BaO(s) +3Cu(s) + ([ 7/2]−x)H2O(g)

35.487g of solid YBa2Cu3O(7−x) reacts completely according to the above equation, in a stream of H2 gas at 1000°C, leaving 32.825g of solid residue.

Molar mass of YBa2Cu3O(7−x) is (666.194 - 15.9994x)
Mass equivalent of (0.5Y2O3 + 2BaO + 3Cu) is 610.196

Find the value of x

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Your molar mass and mass equivalents are "grams per mole" or g/mol

Using 32.825g and 610.196g/mol find the number of moles of (0.5Y2O3 + 2BaO + 3Cu). Note that you get "1 mole" of this mixture from each mole of YBa2Cu3O(7-x) that you start with. I will call this number of moles "y moles" -- you can do the arithmetic.

Since your reactant to product ratio is 1:1,

y mol (product) = 35.487g / (666.194-15.994x)g/mol (reactant)

Since you already calculated y (above), you should now be able to solve this equation for x.
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