What volume of a 2.5 M NaOH solution is required to prepare 2.0 liters of a 0.75M NaOH? A. 2.5 liters B. 6.7 liters C. 0.60 lite
1 answer:
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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=?
- T= 295 K
Replacing in the equation for ideal gas:
1.10 atm* 26.5 L= n* *295 K
Solving:
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 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 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>