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Aloiza [94]
3 years ago
13

PLEASE HELP!! 15POINTS!!! And brainliest

Chemistry
1 answer:
Elis [28]3 years ago
8 0

Answer: a. 3.36 L

b. 33.2 g

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number 6.023\times 10^{23} of particles.

Standard condition of temperature (STP)  is 273 K and atmospheric pressure is 1 atm respectively.

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

\text{Number of moles of} Fe_2O_3=\frac{16.0g}{159.69g/mol}=0.1mole

3O_2(g)+4Fe(s)\rightarrow 2Fe_2O_3(s)

a. 2 moles of Fe_2O_3 are produced by = 3\times 22.4L=67.2L of O_2

Thus 0.1 moles of  Fe_2O_3[tex] are produced by =[tex]\frac{67.2}{2}\times 0.1=3.36L of O_2

b. 3\times 22.4L=67.2L of O_2 react with = 4\times 55.8=223.2g of iron

Thus 10.0 L of O_2 react with = \frac{223.2}{67.2}\times 10=33.2g of iron

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Explanation:

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MnO₄⁻(aq) + I⁻(aq) → Mn²⁺(aq) + I₂(s)

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Step 1: Identify both half-reactions

Reduction: MnO₄⁻(aq) → Mn²⁺(aq)

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Step 2: Perform the mass balance, adding H⁺(aq) and H₂O(l) where appropriate

MnO₄⁻(aq) + 8 H⁺(aq) → Mn²⁺(aq) + 4 H₂O(l)

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Step 3: Perform the charge balance, adding electrons where appropriate

MnO₄⁻(aq) + 8 H⁺(aq) + 5 e⁻ → Mn²⁺(aq) + 4 H₂O(l)

2 I⁻(aq) → I₂(s)  + 2 e⁻

Step 4: Multiply both half-reactions by numbers so that the number of electrons gained and lost are equal

2 × (MnO₄⁻(aq) + 8 H⁺(aq) + 5 e⁻ → Mn²⁺(aq) + 4 H₂O(l))

5 × (2 I⁻(aq) → I₂(s)  + 2 e⁻)

Step 5: Add both half-reactions and cancel what is repeated on both sides

2 MnO₄⁻(aq) + 16 H⁺(aq) + 10 e⁻ + 10 I⁻(aq) → 2 Mn²⁺(aq) + 8 H₂O(l) + 5 I₂(s)  + 10 e⁻

The balanced reaction is:

2 MnO₄⁻(aq) + 16 H⁺(aq) + 10 I⁻(aq) → 2 Mn²⁺(aq) + 8 H₂O(l) + 5 I₂(s)

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To solve this problem, we will use the Boyle's Law, which describes how pressure changes when volume changes and vice-versa. The equation for this law is the following one, and we'll clear for V2:

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Read 2 more answers
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