A gas mixture contains 1.61 moles of hydrogen and 2.31 moles of oxygen. What is the mole fraction of oxygen?

Consider a situation in which 211 g

Stella [2.4K]

Answer:

3.00 mol

Explanation:

Given data:

Mass of P₄ = 211 g

Mass of oxygen = 240 g

Moles of P₂O₅ = ?

Solution:

Chemical equation:

P₄ + 5O₂ → 2P₂O₅

Number of moles of P₄:

Number of moles = mass/ molar mass

Number of moles = 211 g / 123.88 g/mol

Number of moles = 1.7 mol

Number of moles of O₂ :

Number of moles = mass/ molar mass

Number of moles = 240 g / 32g/mol

Number of moles = 7.5 mol

Now we will compare the moles of product with reactant.

O₂ : P₂O₅

5 : 2

7.5 : 2/5×7.5 = 3.00

P₄ : P₂O₅

1 : 2

1.7 : 2×1.7 = 3.4 mol

Oxygen is limiting reactant so the number of moles of P₂O₅ are 3.00 mol.

Mass of P₂O₅:

Mass = number of moles × molar mass

Mass = 3 mol ×283.9 g/mol

Mass = 852 g

3 0

3 years ago

How many grams of NO can be produced if 204 g of NO2 is mixed with 58.1 g of H2O?

Goshia [24]

Answer:

44.4 grams of NO can be produced

Explanation:

Step 1: Data given

Mass of NO2 = 204 grams

Molar mass NO2 = 46.0 g/mol

Mass of H2O = 58.1 grams

Molar mass H2O = 18.02 g/mol

Step 2: The balanced equation

3NO2 + H2O→ 2HNO3 + NO

Step 3: Calculate moles NO2

Moles NO2 = 204 grams / 46.0 g/mol

Moles NO2 = 4.43 moles

Step 4: Calculate moles H2O

Moles H2O = 58.1 grams / 18.02 g/mol

Moles H2O = 3.22 moles

Step 5: Calculate limiting reactant

For 3 moles NO2 we need 1 mol H2O to produce 2 moles HNO3 and 1 mol NO

NO2 is the limiting reactant. It will completely be consumed (4.43 moles). H2O is in excess. there will react 4.43 /3 = 1.48 moles. There will remain 3.22 - 1.48 = 1.74 moles

Step 6: Calculate moles NO

For 3 moles NO2 we need 1 mol H2O to produce 2 moles HNO3 and 1 mol NO

For 4.43 moles NO2 we'll have 4.43/3 = 1.48 moles NO

Step 7: Calculate mass NO

Mass NO = 1.48 moles * 30.01 g/mol

Mass NO = 44.4 grams

44.4 grams of NO can be produced

3 0

3 years ago

A 0.595 g sample of a metal, M, reacts completely with sulfuric acid according to the reaction M ( s ) + H 2 SO 4 ( aq ) ⟶ MSO 4

zalisa [80]

Answer:

molar mass M(s) = 65.326 g/mol

Explanation:

M(s) + H2SO4(aq) → MSO4(aq) + H2(g)

∴ VH2(g) = 231 mL = 0.231 L

∴ P atm = 1.0079 bar

∴ PvH2O(25°C) = 0.03167 bar

Graham´s law:

⇒ PH2(g) = P atm - PvH2O(25°C)

⇒ PH2(g) = 1.0079 bar - 0.03167 bar = 0.97623 bar = 0.9635 atm

∴ nH2(g) = PV/RT

⇒ nH2(g) = ((0.9635 atm)(0.231 L))/((0.082 atmL/Kmol)(298 K))

⇒ nH2(g) = 9.1082 E-3 mol

⇒ n M(s) = ( 9.1082 E-3 mol H2(g) )(mol M(s)/mol H2(g))

⇒ n M(s) = 9.1082 E-3 mol

∴ molar mass M(s) [=] g/mol

⇒ molar mass M(s) = (0.595 g) / (9.1082 E-3 mol)

⇒ molar mass M(s) = 65.326 g/mol

7 0

3 years ago

I need help on my chemistry quiz

slamgirl [31]

Answer:

it could be 25degrees c... not sure

4 0

3 years ago

A mixture of 1.20 mols He, 2.40 mols Ne, 4.80 mols Kr, and 0.60 mols Ar has a total pressure of 600.0 mm Hg. What is the partial

Juliette [100K]

Answer: The partial pressure of the Kr is 320 mm Hg.

Explanation:

According to Raoult's Law , the partial pressure of each component in the solution is equal to the total pressure multiplied by its mole fraction. It is mathematically expressed as

$p_A=x_A\times P_{total}$