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2 years ago

What is the mass of H2O produced when 10.0 grams of H2 reacts completely with 80.0 grams of O2?

1 answer:

8 0

H₂:

M=2g/mol
m=10g

n = m/M = 10g/2g/mol = 5mol

O₂:

M = 32g/mol
m = 80g

n = m/M = 80g/32g/mol = 2,5mol

2H₂ + O₂ ⇒ 2H₂O
2mol : 1mol : 2mol
5mol : 2,5mol : 5mol
10g : 80g : 10g+80g=90g

B

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2 years ago

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jarptica [38.1K]

Answer:

a. Too close to zero

b. ΔS > 0

c. Too close to zero

d. ΔS < 0

e. ΔS > 0

Explanation:

In order to answer this question we need to compare the change in number of mol in the gas state for the products vs the reactants.

An increase in number moles gas leads to a positive change in entropy. Conversely a decrease will mean ΔS is negative.

Now we can solve the parts in this question.

a. H₂ (g) + F₂ (g) ⇒ 2 HF (g)

We have no change in the number of moles gas products minus reactants. Therefore is too close to zero to decide since there is no change in mol gas.

b. 2 SO₃ ( g) ⇒ 2 SO₂ (g) + O₂ (g)

We have three moles of gas products starting with 2 mol gas SO₂, therefore ΔS is positive.

c. CH₄ (g) + 2 O₂ (g) ⇒ CO₂ (g) + 2 H₂O (g)

Again, we have the same number of moles gas in the products and the reactants (3), and it is too close to zero to decide

d. 2 H₂S (g) + 3 O₂(g) ⇒ 2 H2O (g) + 2 SO₂ (g)

Here the change in number of moles gas is negative ( -1 ), so will expect ΔS < 0

e. 2H₂O₂ (l) ⇒ 2H₂O(l) +- O2(g)

Here we have 1 mol gas and 2 mol liquid produced from 2 mol liquid reactants, thus the change in entropy is positive.

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3 years ago

7 f Find the volume in dm3 and in mole of 0.505m of NaoH required to react with 40ml of 0.505m

Anna [14]

The volume of NaOH required is 0.08 dm³

To solve this question, we'll begin by writing the balanced equation for the reaction between H₂SO₄ and NaOH. This is illustrated below:

H₂SO₄ + 2NaOH —> Na₂SO₄ + 2H₂O

From the balanced equation above,

Mole ratio of the acid, H₂SO₄ $(n_{A}) = 1$

Mole ratio of the base, NaOH $(n_{B}) = 2$

Next, we shall determine the volume of NaOH required to react with H₂SO₄. This can be obtained as follow:

Molarity of the base, NaOH $(M_{B}) = 0.505 M$

Volume of the acid, H₂SO₄ $(V_{A}) = 40 mL$

Molarity of the acid, H₂SO₄ $(M_{A}) = 0.505 M$

<h3>Volume of the base, NaOH $(V_{B})$ =? </h3>

$\frac{M_{A} * V_{A}}{M_{B} * V_{B}} = \frac{n_{A}}{n_{B}}\\\\\frac{0.505 * 40}{0.505 *V_{B}} = \frac{1}{2}\\\\\frac{20.2}{0.505 *V_{B}} = \frac{1}{2}$