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3 years ago

What is the molar mass of a gas with a density of 3.50 g/L at a

Chemistry

1 answer:

Kruka [31]3 years ago

3 0

Answer:

104 g/mol

Explanation:

Use the ideal gas equation:

$pV=nRT$

Where:

p is pressure: 0.950atm
V is volume: unknown
n is number of moles: unknown
R is the universal constat of gases: 0.08206 atm.liter/ (K.mol)
T is the absolute temperature: 345K

Use the molar mass of the gas to include the density in the formula:

molar mass = mass in grams / number of moles

⇒ mass in grams = number of moles × molar mass

density = mass in grams / volume

⇒ density = number of moles × molar mass / volume

density = (n/V) × molar mass

⇒ n/V = density / molar mass

Clear n/V from the gas ideal equation and subsittute with density/molar mass:

n/V = p/(RT)

density / molar mass = n/V

density/molar mass = p/(RT)

molar mass = density × RT / p

Now you can subsitute the data:

molar mass = (3.50g/liter) × 0.08206 atm.liter/(K.mol) × 345K / 0.950 atm

molar mass = 104.3 g/mol

Round to the nearest whole number: 104g/mol ← answer

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Use this equation for the following problems: 2NaN3 --> 2Na+3N2

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Answer:

1) 65.0

2) 16.434 L = 16434 mL.

Explanation:

2NaN₃ → 2Na + 3N₂,

It is clear from the balanced equation that 2.0 moles of NaN₃ are decomposed to 2.0 moles of Na and 3.0 moles of N₂.

Q1: How many grams of NaN₃ are needed to make 23.6L of N₂?

Density of N₂ = 0.92 g/L which means that every 1.0 L of N₂ contains 0.92 g of N₂.

Now, we can get the mass of N₂ in 23.6 L N₂ using cross multiplication:

1.0 L of N₂ contains → 0.92 g of N₂.

23.6 L of N₂ contains → ??? g of N₂.

∴ The mass of N₂ in 23.6 L of N₂ = (23.6 L)(0.92 g)/(1.0 L) = 21.712 g.

We can get the no. of moles of 23.6 L of N₂ (21.712 g) using the relation:

n = mass/molar mass = (21.712 g)/(28.0 g/mol) = 0.775 mol.

We can get the no. of moles of NaN₃ needed to produce 0.775 mol of N₂:

using cross multiplication:

2.0 moles of NaN₃ produce → 3.0 moles of N₂, from the balanced equation.

??? mol of NaN₃ produce → 0.775 moles of N₂.

∴ The no. of moles of NaN₃ needed = (2.0 mol)(0.775 mol)/(3.0 mol) = 0.517 mol.

Finally, we can get the grams of NaN₃ needed:

mass = no. of moles x molar mass = (0.517 mol)(65.0 g/mol) = 33.6 g.

Q2: How many mL of N₂ result if 8.3 g Na are also produced?

We need to get the no. of moles of 8.3 g Na using the relation:

n = mass/atomic mass = (8.3 g)/(22.98 g/mol) = 0.36 mol.

We can get the no. of moles of N₂ produced with 0.36 mol of Na:

using cross multiplication:

2.0 moles of Na produced with → 3.0 moles of N₂, from the balanced equation.

0.36 moles of Na produced with → ??? moles of N₂.

∴ The no. of moles of N₂ needed = (3.0 mol)(0.36 mol)/(2.0 mol) = 0.54 mol.

We can get the mass of 0.54 mol of N₂:

mass = no. of moles x molar mass = (0.54 mol)(28.0 g/mol) = 15.12 g.

Now, we can get the mL of 15.12 g of N₂:

using cross multiplication:

1.0 L of N₂ contains → 0.92 g of N₂, from density of N₂ = 0.92 g/L.

??? L of N₂ contains → 15.12 g of N₂.

∴ The volume of N₂ result = (1.0 L)(15.12 g)/(0.92 g) = 16.434 L = 16434 mL.

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