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allochka39001 [22]
3 years ago
6

The molar mass of H2O is 18.01 g/mol. The molar mass of O2 is 32.00 g/mol. What mass of H2O, in grams, must react to produce 50.

00 g of O2?
Chemistry
2 answers:
Annette [7]3 years ago
8 0
This reaction is called the electrolysis of water. The balanced reaction is:
2H2O = 2H2 + O2
We are given the amount of O2 produced from the electrolysis reaction. This will be the starting point of our calculation.

50.00 grams O2 ( 1 mol O2 / 32 grams O2) ( 2 mol H2O / 1 mol O2) ( 18.01 g H2O / 1 mol H2O ) = 56.28 g H2O
Tomtit [17]3 years ago
7 0

Answer:

Mass of H2O = 56.27 g

Explanation:

<u>Given:</u>

Molar mass of H2O = 18.00 g/mol

Molar mass of O2 = 32.00 g/mol

Mass of O2 produced = 50.00 g

<u>To determine:</u>

Mass of H2O reacted

<u>Explanation:</u>

Water breaks apart during electrolysis to produce O2 and H2. This can be represented as:

2H2O \rightarrow 2H2 + O2

Moles of O2 produced = \frac{Mass\ of\ O2}{Molar\ Mass} = \frac{50.00}{32.00} = 1.563

Based on the reaction stoichiometry:

2 moles of H2O produces 1 mole of O2

Therefore, moles of H2O that would react to produce 1.563 moles of O2 is:

= \frac{1.563\ moles\ O2*2\ moles\ H2O}{1\ mole\ O2} =3.126 moles

Mass of H2O = Moles of H2O * Molar mass \\= 3.126 moles * 18 g/mol = 56.27 \ g

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chubhunter [2.5K]

Answer:

A.

Explanation:

Water was added to the reaction after the completion of the reaction so as to lower the solubility if the product in the solution therefore, the product can be precipitated out. On adding water the reaction moves in forward direction and more product is formed. (By Le Chatelier's principle). Thus, the precipitation occurs. Hence, option A is correct.

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A decomposition reaction, with a rate that is observed to slow down as the reaction proceeds, has a half-life that does not depe
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Since it was explicitly stated in the question that the half life is independent of the initial concentration of the reactant then the third option must necessarily be false. Also, the plot of the natural logarithm of the concentration of reactant against time for a first order reaction is linear. In a first order reaction, the half life is independent of the initial concentration of the reactant. Hence the answer.

3 0
3 years ago
A compound with a molar mass of 60g/mol is 40.4% carbon, 6.7% hydrogen and 53.3% oxygen (by mass). determine the emperical and m
Fittoniya [83]

<span>A compound is found to be 40.0% carbon, 6.7% hydrogen and 53.5% oxygen. Its molecular mass is 60. g/mol. 
</span>Q1)
Empirical formula is the simplest ratio of whole numbers of components making up a compound.
the percentages have been given, therefore we can calculate for 100 g of the compound.

                                C                            H                        O
Mass in 100 g      40.0 g                       6.7 g                   53.5 g
Molar mass            12 g/mol                1 g/mol                 16 g/mol
Number of moles   40.0/12= 3.33         6.7/1 = 6.7          53.5/16 = 3.34
Divide by the least number of moles  
                             3.33/3.33 = 1           6.7/3.33 = 2.01   3.34/3.33 = 1.00
after rounding off
C - 1 
H - 2
O - 1

Empirical formula - CH₂O

Q2)
Molecular formula is the actual number of components making up the compound.
To find the number of empirical units we have to find the mass of one empirical unit.
Mass of one empirical unit = CH₂O - 12 + (1x2) + 16 = 30 g
Mass of one mole of compound = 60 g
Number of empirical units = 60 g / 30 g = 2
Therefore molecular formula - 2(CH₂O) 
 Molecular formula - C₂H₄O₂
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Answer:

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