Practice_EXAM_I

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Unformatted text preview: ide, which leads us to put 2 (called coefficient) in front of the PH3. Now the equation is updated to 2 PH3 + __O2 1 P2O5 + __H2O As there are 6 hydrogen atoms in 2 PH3, thus we need to balance the hydrogen atoms at the right side, which lead us to put 3 in front of the H2O. Now the equation is updated to 2 PH3 + __O2 1 P2O5 + 3 H2O Now we need to balance the oxygen atoms. Since there are 1x5+3x1 = 8 oxygen atoms at the right side, and thus the left side must have the same number of oxygen atoms. That is to say, __x2 = 8. So __ = 4. Now the equation is updated to 2 PH3 + 4 O2 1 P2O5 + 3 H2O Since all the atoms of each type has the same amount, this equation is balanced. 22. In the reaction of Al(OH)3 with H2SO4, how many moles of water can be produced If the reaction is begun with 5.500 mole of Al(OH)3? 2Al(OH)3 + 3H2SO4 Al2(SO4)3 + 6H2O (a) 2.50 (b) 4.75 (c) 6.32 (d) 7.58 (e) 16.50 Hint: For 11th ed.: p.p. 95‐97 and Examples 3.13 and 3.14 (application of Figure 3.8); for 9th edition: p.p. 97‐101. Examples 3.13. and 3.1...
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This note was uploaded on 01/25/2014 for the course CHEM 1411 taught by Professor None during the Fall '11 term at HCCS.

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