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ki77a [65]
3 years ago
11

For the reaction ?H2 +?O2 → ?H2O, what is the maximum amount of H2O (18.0153 g/mol) which could be formed from 9.43 g of H2 (2.0

1588 g/mol) and 12.98 g of O2 (31.9988 g/mol)? Answer in units of g.
Chemistry
1 answer:
Montano1993 [528]3 years ago
3 0
2H2 + O2 →2H2O
This is a limiting reactant problem in which you need to figure out which reactant is the limiting reactant. You can usually look at the information given and see that hydrogen has the smaller amount of mass used, but it is always good to check that assumption through stoichiometry:
(9.43g H2/2.02g/mol)(2mol H2O/2mol H2)(18.02gH2O/1mol H2O)= 84.1g H2O produced from 9.43g of H2

(12.98g O2/32.00g/mol O2)(2mol H2O/1mol O2)(18.02gH2O/1mol H2O)= 14.6g H2O

So my assumption that hydrogen makes the smaller amount of water is wrong since oxygen (it is the limiting reactant) created the smaller amount of water with 14.6g, so 14.6g is the maximum amount of water that can be formed
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The question is incomplete, the complete reaction equation is;

The concentration of a hydrogen peroxide solution can be determined by titration

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2.275 M

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The equation of the reaction is;

2 MnO4^-(aq) + 16 H^+(aq) + 5H2O2(aq) -------> 2Mn^+(aq) + 10H^+ (aq) + 8H2O(l)

Let;

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VA = volume of MnO4^- = 24.35 cm3

VB = volume of H2O2 = 25.0 cm3

NA = number of moles of  MnO4^- = 2

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CAVA/CBVB = NA/NB

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C1V1 = C2V2

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V1 = initial volume of H2O2 solution =  5.0 cm3

C2 = final concentration of H2O2 solution= 0.0455 M

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C1 = 0.0455 * 250/5

C1 = 2.275 M

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