eview Topics Use the References to access important values if needed for this qu

Question

eview Topics Use the References to access important values if needed for this question. For the following reaction, 31.1 grams of zinc oxide are allowed to react with 6.09 grams of water zinc oxide (s) + water()-> zinc hydroxide (aq) what is the maximum amount of zinc hydroxide that can be formed? L grams 1 | What is the FORMULA for the limiting reagent? what amount of the excess reagent remains after the reaction is complete? grams Submit Answer Retry Entire Group 8 more group attempts remaining

Explanation / Answer

mol of ZnO = mass/MW = 31.1/81.408 = 0.3820 mol

mol of water = mass/MW = 6.09/18 = 0.3383 mol

raito is

ZnO + H2O = Zn(OH)2

then

ratio is 1:1

there is less water than ZnO, then

wate rlimits reaction

max Zn(OH)2 from water:

1 mol of Zn(OH)2 = 1 mol of water

0.3383 mol of water --> 0.3383 mol of Zn(OH)2

mass = mol*MW = (0.3383 )(99.3947) = 33.625 g of Zn(OH)2

b)

limiting reagent is water, so H2O(l)

c)

excess reagent is ZnO

mol left = 0.382-0.3383 = 0.0437

mass ZnO = mol*MW = 0.0437*81.408 = 3.5575 g of ZnO left

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