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kvasek [131]
3 years ago
11

Solid sodium azide (NaN3) produces solid sodium and nitrogen gas. How many grams of sodium azide are needed to yield a volume of

26.5 L of nitrogen gas at a temperature of 295 K and a pressure of 1.10 atmospheres
Chemistry
2 answers:
ch4aika [34]3 years ago
6 0

Answer:

52.008 grams of sodium azide are needed to yield a volume of 26.5 L of nitrogen gas at a temperature of 295 K and a pressure of 1.10 atmospheres.

Explanation:

An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of the gases:

P*V = n*R*T

In this case, the balanced reaction is:

2 NaN₃ → 2 Na + 3 N₂

You know the following about N₂:

  • P= 1.10 atm
  • V= 26.5 L
  • n=?
  • R=0.082057 \frac{atm*L}{mol*K}
  • T= 295 K

Replacing in the equation for ideal gas:

1.10 atm* 26.5 L= n* 0.082057 \frac{atm*L}{mol*K}*295 K

Solving:

n=\frac{1.10 atm*26.5 L}{0.082057 \frac{atm*L}{mol*K} *295K}

n= 1.2 moles

Now, the following rule of three can be applied: if 3 moles of N₂ are produced by stoichiometry of the reaction from 2 moles of NaN₃, 1.2 moles of N₂ are produced from how many moles of NaN₃?

moles of NaN_{3}=\frac{1.2 molesofN_{2} *2 molesofNaN_{3} }{3 molesofN_{2} }

moles of NaN₃= 0.8

Since the molar mass of sodium azide is 65.01 g / mol, then one last rule of three applies: if 1 mol has 65.01 grams of NaN₃, 0.8 mol how much mass does it have?

mass of NaN_{3} =\frac{0.8 mol*65.01 grams}{1 mol}

mass of NaN₃=52.008 grams

<u><em>52.008 grams of sodium azide are needed to yield a volume of 26.5 L of nitrogen gas at a temperature of 295 K and a pressure of 1.10 atmospheres.</em></u>

fredd [130]3 years ago
6 0

Answer:

We need 51.7 grams of NaN3

Explanation:

Step 1 : Data given

Volume of N2 gas produced = 26.5 L

Temperature = 295 K

Pressure = 1.10 atm

Step 2: The balanced equation

2NaN3 → 2Na + 3N2

Step 3: Calculate moles N2

p*V = n*R*T

⇒with p = the pressure of nitrogen gas = 1.10 atm

⇒with V = volume of nitrogen gas = 26.5 L

⇒with n = the number of moles nitrogen gas = TO BE DETERMINED

⇒with R = the gas constant = 0.08206 L*atm/mol* K

⇒with T = the themperature = 295

n = (p*V)/(R*T)

n = (1.10 * 26.5) / (0.08206 * 298)

n = 1.192 moles N2

Step 4: Calculate moles NaN3

For 2 moles NaN3 we'll have 2 moles Na and 3 moles N2

For 1.192 moles N2 we need 2/3 * 1.192  = 0.795 moles NaN3

Step 5: calculate mass NaN3

Mass NaN3 = moles NaN3 * molar mass NaN3

Mass NaN3 = 0.795 moles * 65.0 g/mol

Mass NaN3 = 51.7 gtams NaN3

We need 51.7 grams of NaN3

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rusak2 [61]

Answer:

60.42% is the percent yield of the reaction.

Explanation:

Moles of methane gas at 734 Torr and a temperature of 25 °C.

Volume of methane gas = V = 26.0 L

Pressure of the methane gas = P = 734 Torr = 0.9542 atm

Temperature of the methane gas = T = 25 °C = 298.15 K

Moles of methane gas = n

PV=nRT

n=\frac{PV}{RT}=\frac{0.9542 atm\times 26.0L}{0.0821 atm L/mol K\times 298.15 K}=1.0135 mol

Moles of water vapors at 700 Torr and a temperature of 125 °C.

Volume of water vapor = V' = 23.0 L

Pressure of water vapor = P' = 700 Torr = 0.9100 atm

Temperature of  water vapor = T' = 125 °C = 398.15 K

Moles of water vapor gas = n'

P'V'=n'RT'

n'=\frac{PV}{RT}=\frac{0.9100 atm\times 23.0L}{0.0821 atm L/mol K\times 398.15 K}=0.6402 mol

CH_4(g)+H_2O(g)\rightarrow CO(g)+3H_2(g)

According to reaction , 1 mol of methane reacts with 1 mol of water vapor. As we can see that moles of water vapors are in lessor amount which means it is a limiting reagent and formation of hydrogen gas will depend upon moles of water vapors.

According to reaction 1 mol of water vapor gives 3 moles of hydrogen gas.

Then 0.6402 moles of water vapor will give:

\frac{3}{1}\times 0.6402 mol=1.9208 mol of hydrogen gas

Moles of hydrogen gas obtained theoretically = 1.9208 mol

The reaction produces 26.0 L of hydrogen gas measured at STP.

At STP, 1 mole of gas occupies 22.4 L of volume.

Then 26 L of volume of gas will be occupied by:

\frac{1}{22.4 L}\times 26 L= 1.1607 mol

Moles of hydrogen gas obtained experimentally = 1.1607 mol

Percentage yield of hydrogen gas of the reaction:

\frac{Experimental}{Theoretical}\times 100

\%=\frac{ 1.1607 mol}{1.9208 mol}\times 100=60.42\%

60.42% is the percent yield of the reaction.

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Answer:

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Explanation:

The group 7A elements consists of the most reactive non-metals on the periodic table.

This group is known as the group of halogens. They consist of element fluorine, chlorine, bromine, iodine and astatine.

  • The elements in this group have the highest electronegativity values.
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Answer:

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what is the molecular formula of the compound?

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Explanation:

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Bromine has one more electron shell than the chlorine atom making the radius larger than the chlorine atom.
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