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Feliz [49]
2 years ago
13

How many moles of Al will be consumed when 0.400 mol of Al2O3 are produced in

Chemistry
1 answer:
Vladimir [108]2 years ago
7 0

Answer:

C) 0.800 mol

Explanation:

  • 4Al + 3O₂ → 2Al₂O₃

In order to <u>convert from moles of Al₂O₃ into moles of Al</u>, we'll need to use<em> the stoichiometric coefficients</em>, using a conversion factor that has Al₂O₃ moles in the denominator and Al moles in the numerator:

  • 0.400 mol Al₂O₃ * \frac{4molAl}{2molAl_2O_3} = 0.800 mol Al

So the correct answer is option C).

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Ion

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When 10.0 grams of sulfur reacts with fluorine gas at a pressure of 2.69 atmosphere in a 5.00 L container at 0.00 degrees Celsiu
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Answer:

74.1%

Explanation:

Based on the reaction:

S₈ + 16F₂ → 8SF₄

<em>1 mole of sulfur reacts with 16 moles of F₂ to produce 8 moles of SF₄</em>

<em />

To solve this question we must find the moles of each reactant in order to find the moles of SF₄. Thus, we can find the theoretical mass produced. Percent yield is:

Percent yield = Actual yield (25.0g) / Theoretical yield * 100

<em>Moles S₈: 256.52g/mol</em>

10.0g * (1mol / 256.52g) = 0.0390 moles

<em>Moles F₂:</em>

<em>PV = nRT</em>

PV/RT = n

<em>Where P is pressure in atm, V is volume in liters, R is gas constant and T is absolute temperature (0°C = 273.15K)</em>

2.69atm*5.00L / 0.082atmL/molK*273.15K = n

0.600 moles = n

For a complete reaction of 0.600 moles F₂ are required:

0.600mol F₂ * (1mol S₈ / 8 mol F₂) = 0.075 moles S₈

As there are just 0.0390 moles, S₈ is limiting reactant.

The theoretical moles and mass of SF₄ -Molar mass: 108.07g/mol- is:

0.0390 moles S₈ * (8mol SF₄ / 1mol S₈) = 0.312 moles SF₄ * (108.07g) =

33.7g

Percent yield = 25.0g / 33.7g * 100

= 74.1%

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2 years ago
What is the Law of Definite Composition and how does it apply to pure substances and mixtures?
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The Law of Definite Proportions ensures that chemical compounds are always created using the same proportions, regardless of the amount of the compound which is being made

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A sample compound with a molar mass of 34.00g/mol is found to consist of 0.44g H and 6.92g O. Calculate both empirical and molec
balu736 [363]

Answer:

E.F= OH

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Explanation:

Empirical Formula

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*note leave extra sig figs for calculations

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Step 2: Identify which is the smallest mol

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Step 4: use calculations as subscripts

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Molecular Formula:

Step 1: identify  molecular mass and mass from the E.F

the molecular mass is given 34.00g/mol

the mass of the E.F is

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Step 2:Divide the molecular mass by the mass given by the emipirical formula.

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Step 3:Multiply the empirical formula (the subscripts) by this number to get the molecular formula.  ANSWER: M.F=2(OH)- O_{2} H_{2}

3 0
2 years ago
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