Question
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
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
Anonymous
the above is wrong, its 2*16 not 2+16