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Zinaida [17]
3 years ago
12

How many water molecules (H2O) can be produced from 6 molecules of hydrogen gas (white) reacting with 6 molecules of oxygen gas

(red)? Which reactant is the limiting reactant?
Chemistry
2 answers:
Vladimir79 [104]3 years ago
5 0
I think you can only have 3 water molecules because you need 2 hydrogen molecules in every water molecule and you have 6 hydrogen molecules so 6/2=3 and the reactant that is limited would be hydrogen since it limits the amount of water molecules you can have
Elza [17]3 years ago
3 0

<u>Answer:</u> The hydrogen gas is the limiting reagent and 6 molecules of water are produced.

<u>Explanation:</u>

We are given:

Molecules of hydrogen gas = 6 molecules

Molecules of oxygen gas = 6 molecules

The chemical equation for the reaction of hydrogen gas and oxygen gas follows:

2H_2+O_2\rightarrow 2H_2O

By Stoichiometry of the reaction:

2 molecules of hydrogen gas reacts with 1 molecule of oxygen gas

So, 6 molecules of hydrogen gas will react with = \frac{1}{2}\times 6=3 molecules of oxygen gas

As, given amount of oxygen gas is more than the required amount. So, it is considered as an excess reagent.

Thus, hydrogen gas is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

2 molecules of hydrogen gas produces 2 molecules of water

So, 6 molecules of hydrogen gas will produce = \frac{2}{2}\times 6=6 molecules of water

Hence, the hydrogen gas is the limiting reagent and 6 molecules of water are produced.

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Find the pH of the equivalence point and the volume (mL) of 0.0372 M NaOH needed to reach the equivalence point in the titration
elixir [45]

Answer:

8.54

Explanation:

At equivalence point :  

42.2 X 0.052 = Vol. NaOH X 0.0372

Vol of NaOH = 2.1944/0.0372 = 58.99 ml

So volume of NaOH recquired to reach equivalence point = 58.99 ml

Number of miliimoles of CH3COOH = molarity X volume in ml = 42.2 X 0.052             = 2.1944 millimoles

Number of millimoles of NaOH = 58.99 X 0.0372 = 2.1944

Now CH₃COOH and NaOH reacts to give CH₃COONa according to the reaction :

CH₃COOH + NaOH ------> CH₃COONa + H₂O

1 mole of CH₃COOH reacts with 1 mole of NaOH to give 1 mole of CH₃COONa  

So 2.1944 millimoles of CH₃COOH will react with 2.1944 millimoles of NaOH to give 2.1944 millimoles of CH₃COONa

So all the acid (CH₃COOH) and base (NaOH) has been converted into salt (CH₃COONa) so there is no acid or base left.

Now molarity of CH₃COONa = number of millimoles of CH₃COONa/total volume in ml = 2.1944/(58.99 + 42.2) = 2.1944/101.19 = 0.02169 M

So using the hydrolysis equation :  

pH = 1/2 [ pKw + pKa + log c ]  

Ka for acetic acid = 1.75 X 10⁻⁵  

so pKa = -log (1.75 X 10⁻⁵) = 4.74  

Kw = 10⁻¹⁴

so pKw = -log 10⁻¹⁴ = 14

c = 0.02169  

so log c = log 0.02169 = -1.66  

putting the values....  

pH = 1/2 [14 + 4.74 - 1.66 ]  

pH = 1/2 [ 17.08] = 8.54

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