Asked by Anonymous
According to the following reaction, how many grams of zinc hydroxide will be formed upon the complete reaction of 29.8 grams of water with excess zinc oxide?
ZnO (s) + H2O (l) Zn(OH)2 (aq)
ZnO (s) + H2O (l) Zn(OH)2 (aq)
Answers
Answered by
Damon
ZnO (s) + H2O (l) ---> Zn(OH)2 (aq)
same number of moles of Zn(OH)2 as of water
water = 2 + 16 = 18 grams/mol
so
29.8/18 = 1.66 moles of H2O and of Zn(OH)2
Zn = 65
2O = 32
2H = 2
-------
99 grams/mole
99*1.66 = 154 grams Zn(oh)2
same number of moles of Zn(OH)2 as of water
water = 2 + 16 = 18 grams/mol
so
29.8/18 = 1.66 moles of H2O and of Zn(OH)2
Zn = 65
2O = 32
2H = 2
-------
99 grams/mole
99*1.66 = 154 grams Zn(oh)2
Answered by
Anonymous
the above is wrong, its 2*16 not 2+16
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