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3 years ago

If 4.0 g of hydrogen and 10.0 g of oxygen are mixed according to the equation 2 h2 + o2 → 2 h2o, which is the limiting reagent?

1 answer:

8 0

From the equation, we can see that the molar ratio between hydrogen and oxygen is:
2 : 1

Next, we determine the moles of hydrogen and oxygen that are actually present using:
moles = mass / Mr

Hydrogen:
moles = 4 / 2 = 2

Oxygen:
10/32 = 0.3125

Therefore, it is evident that the moles of oxygen present, 0.3125, are less than those that are required for 2 moles of hydrogen, which is 1. This makes oxygen the limiting reactant, which is the one that limits the completion of a reaction.

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2 years ago

Be sure to answer all parts. What is the [H3O+] and the pH of a buffer that consists of 0.26 M HNO2 and 0.89 M KNO2? (K, of HNO2

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Answer : The $H_3O^+$ ion concentration is, $1.12\times 10^{-3}M$ and the pH of a buffer is, 2.95

Explanation : Given,

$K_a=7.1\times 10^{-4}$

Concentration of $HNO_2$ (weak acid)= 0.26 M

Concentration of $KNO_2$ (conjugate base or salt)= 0.89 M

First we have to calculate the value of $pK_a$ .

The expression used for the calculation of $pK_a$ is,

$pK_a=-\log (K_a)$

Now put the value of $K_a$ in this expression, we get:

$pK_a=-\log (7.1\times 10^{-4})$

$pK_a=4-\log (7.1)$

$pK_a=3.15$

Now we have to calculate the pH of the solution.

Using Henderson Hesselbach equation :

$pH=pK_a+\log \frac{[Salt]}{[Acid]}$

$pH=pK_a+\log \frac{[KNO_2]}{[HNO_2]}$

Now put all the given values in this expression, we get:

$pH=3.15+\log (\frac{0.89}{0.26})$

$pH=2.95$

The pH of a buffer is, 2.95

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$pH=-\log [H_3O^+]$

$2.95=-\log [H_3O^+]$

$[H_3O^+]=1.12\times 10^{-3}M$

The $H_3O^+$ ion concentration is, $1.12\times 10^{-3}M$

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