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blagie [28]
3 years ago
5

Stoichiometry

Chemistry
1 answer:
cluponka [151]3 years ago
8 0
Stoichiometry is the ratio of reactants to products in a balanced chemical reaction equation. 
the stoichiometry in this case of H₂ to O₂ is 2:1.
This means that 2 moles of H₂ reacts with 1 mol of O₂.
we have been asked to calculate the volume of H₂ gas.
Since both O₂ and H₂ are gases, at standard conditions its stated that molar volume of gases is 22.4 L
This means that at standard temperature and pressure, 1 mol of any gas occupies a volume of 22.4 L
first we need to calculate the number of O₂ moles reacted;
22.4 L of gas - 1 mol of O₂
1 L of gas - 1/22.4 mol/L 
then 20 L of O₂ - 1/22.4 mol/L * 20 L = 0.89 mol
stoichiometry of H₂ to O₂ is 2:1
the number of H₂ moles = 0.89*2 = 1.79 mol
1 mol occupies 22.4 L
Therefore 1.79 mol = 22.4 L/mol * 1.79 mol = 40 L
Therefore it can be seen that stoichiometry applies to volumes as well.
volume of H₂ : O₂ = 40 L : 20 L = 2:1
volume and moles both can be determined by stoichiometry.
Volume of H₂ reacted = 40 L
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Answer: I think it's B.

Explanation:

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2 years ago
A student is asked to standardize a solution of potassium hydroxide. He weighs out 1.08 g potassium hydrogen phthalate (KHC8H4O4
natka813 [3]

Answer:

A. 0.143 M

B. 0.0523 M

Explanation:

A.

Let's consider the neutralization reaction between potassium hydroxide and potassium hydrogen phthalate (KHP).

KOH + KHC₈H₄O₄ → H₂O + K₂C₈H₄O₄

The molar mass of KHP is 204.22 g/mol. The moles corresponding to 1.08 g are:

1.08 g × (1 mol/204.22 g) = 5.28 × 10⁻³ mol

The molar ratio of KOH to KHC₈H₄O₄ is 1:1. The reacting moles of KOH are 5.28 × 10⁻³ moles.

5.28 × 10⁻³ moles of KOH occupy a volume of 36.8 mL. The molarity of the KOH solution is:

M = 5.28 × 10⁻³ mol / 0.0368 L = 0.143 M

B.

Let's consider the neutralization of potassium hydroxide and perchloric acid.

KOH + HClO₄ → KClO₄ + H₂O

When the molar ratio of acid (A) to base (B) is 1:1, we can use the following expression.

M_{A} \times V_{A} = M_{B} \times V_{B}\\M_{A} = \frac{M_{B} \times V_{B}}{V_{A}} \\M_{A} = \frac{0.143 M \times 10.1mL}{27.6mL}\\M_{A} =0.0523 M

8 0
3 years ago
Suppose a 0.025M aqueous solution of sulfuric acid (H2SO4) is prepared. Calculate the equilibrium molarity of SO4−2. You'll find
FromTheMoon [43]

<u>Answer:</u> The concentration of SO_4^{2-} at equilibrium is 0.00608 M

<u>Explanation:</u>

As, sulfuric acid is a strong acid. So, its first dissociation will easily be done as the first dissociation constant is higher than the second dissociation constant.

In the second dissociation, the ions will remain in equilibrium.

We are given:

Concentration of sulfuric acid = 0.025 M

Equation for the first dissociation of sulfuric acid:

       H_2SO_4(aq.)\rightarrow H^+(aq.)+HSO_4^-(aq.)

            0.025          0.025       0.025

Equation for the second dissociation of sulfuric acid:

                    HSO_4^-(aq.)\rightarrow H^+(aq.)+SO_4^{2-}(aq.)

<u>Initial:</u>            0.025            0.025      

<u>At eqllm:</u>      0.025-x          0.025+x        x

The expression of second equilibrium constant equation follows:

Ka_2=\frac{[H^+][SO_4^{2-}]}{[HSO_4^-]}

We know that:

Ka_2\text{ for }H_2SO_4=0.01

Putting values in above equation, we get:

0.01=\frac{(0.025+x)\times x}{(0.025-x)}\\\\x=-0.0411,0.00608

Neglecting the negative value of 'x', because concentration cannot be negative.

So, equilibrium concentration of sulfate ion = x = 0.00608 M

Hence, the concentration of SO_4^{2-} at equilibrium is 0.00608 M

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