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3 years ago

how many molecule of carbon dioxide are needed to react with excess iron oxide to produce 11.6 g of iron

Chemistry

1 answer:

lesya692 [45]3 years ago

5 0

Answer:

0.16 moles of Carbon

Explanation:

The balanced reaction equation:

$2Fe_{2}O_{3}$ + $3C$ → $4Fe$ + $3CO_{2}$ ↑

The mole ratio of Carbon to Iron is 3 : 4 (since Fe2O3 is in excess)

i.e 3 moles of C produces 4 moles of Fe.

If 1 mole of Fe - 55.8g of Fe

? moles - 11.6g of Fe

= $\frac{11.6}{55.8}$ = 0.208 moles

But 3 moles of C - 4 moles of Fe

? moles of C - 0.208 moles of Fe

= $\frac{3 *0.208}{4}$ = 0.16 moles of carbon.

I hope this explanation was clear and useful.

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2 years ago

What is difference in reaction of O2 with metal and nonmetal?

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2 years ago

Given the following data:N2(g) + O2(g)→ 2NO(g), ΔH=+180.7kJ2NO(g) + O2(g)→ 2NO2(g), ΔH=−113.1kJ2N2O(g) → 2N2(g) + O2(g), ΔH=−163

statuscvo [17]

Answer:

ΔH = +155.6 kJ

Explanation:

The Hess' Law states that the enthalpy of the overall reaction is the sum of the enthalpy of the step reactions. To do the addition of the reaction, we first must reorganize them, to disappear with the intermediaries (substances that are not presented in the overall reaction).

If the reaction is inverted, the signal of the enthalpy changes, and if its multiplied by a constant, the enthalpy must be multiplied by the same constant. Thus:

N₂(g) + O₂(g) → 2NO(g) ΔH = +180.7 kJ

2NO(g) + O₂(g) → 2NO₂(g) ΔH = -113.1 kJ

2N₂O(g) → 2N₂(g) + O₂(g) ΔH = -163.2 kJ

The intermediares are N₂ and O₂, thus, reorganizing the reactions:

N₂(g) + O₂(g) → 2NO(g) ΔH = +180.7 kJ

NO₂(g) → NO(g) + (1/2)O₂(g) ΔH = +56.55 kJ (inverted and multiplied by 1/2)

N₂O(g) → N₂(g) + (1/2)O₂(g) ΔH = -81.6 kJ (multiplied by 1/2)

------------------------------------------------------------------------------------

N₂O(g) + NO₂(g) → 3NO(g)

ΔH = +180.7 + 56.55 - 81.6

ΔH = +155.6 kJ

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3 years ago

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FinnZ [79.3K]

Answer:

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Hope this helps!

4 0

2 years ago

Read 2 more answers

Determine the concentrations of mgcl2, mg2 , and cl– in a solution prepared by dissolving 2.75 × 10–4 g mgcl2 in 1.75 l of water

inysia [295]

Answer:

M of MgCl₂ = 1.65 × 10⁻⁶ M

M of Mg²⁺ = 1.65 × 10⁻⁶ M

M of Cl⁻ = 3.30 × 10⁻⁶ M

Explanation:

1) MgCl₂

Molarity = number of moles of solute / volume of solution in liters, M = n / V

n = mass in grams / molar mass

molar mass of MgCl₂ = 24.305 g/mol + 2(35.543 g/mol) = 95.211 g/mol

n = 2.75 × 10⁻⁴ g / 95.211 g/mol = 2.89×10⁻³ moles

⇒ M = n / V = 2.89×10⁻³ moles / 1.75 l = 1.65 × 10⁻⁶ M

2) Mg²⁺ and Cl⁻

Those are the ions in solution.

You assume 100% dissociation of the ionic compound (strong electrolyte).

Then the equation is: MgCl₂ → Mg²⁺ + 2Cl⁻

That means that 1 mol of MgCl₂ produces 1 mol of Mg²⁺ and 2 moles of Cl⁻.

That yields the same molarity concentration of Mg²⁺ , while the molarity concentration of Cl⁻ is the double.

So, the results are:

M of MgCl₂ = 1.65 × 10⁻⁶ M

M of Mg²⁺ = 1.65 × 10⁻⁶ M

M of Cl⁻ = 3.30 × 10⁻⁶ M

8 0

2 years ago

Read 2 more answers