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lyudmila [28]
2 years ago
14

Plz help How many moles are in 3.57 liters of NH3 at STP?

Chemistry
1 answer:
nlexa [21]2 years ago
6 0

<u>1) determine which formula you are going to use.</u>

PV=nRT

<u>2) list your information.</u>

P = 101.325 kPa

V = 3.57 L

n = ?

R = 8.314 \frac{kPa\ L}{mol\ K}

T = 273.15 K

<u>3) rearrange your formula</u>

n=\frac{PV}{RT}

<u>4) solve.</u>

n=\frac{(101.325\ kPa)(3.57\ L)}{(8.314\frac{kPa\ L}{mol\ K})(273.15\ K) } \\\\n=0.1592845319

<em>n = 0.159 mol NH₃</em>

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Answer:

V = 36.7L of N_{2}

Explanation:

1. Write the chemical reaction for the decomposition of sodium azide:

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2. Find the number of moles of N_{2} produced by the decomposition of 71.4g of NaN_{3}:

71.4gNaN_{3}*\frac{1molNaN_{3}}{65gNaN_{3}}*\frac{3molesN_{2}}{2molesNaN_{3}}=1.65molesN_{2}

3. Use the ideal gas equation to find the Volume of N_{2} occupied by 1.65 moles of N_{2}, at the temperature and pressure given by the problem:

PV=nRT

Solving for V:

V=\frac{nRT}{P}

Converting the temperature from ◦C to K:

25◦C+273.15=298.15K

Replacing values:

V=\frac{1.65mol*(0.082\frac{atm.L}{mol.K})*298.15K}{1.10atm}

V=36.7L of N_{2}

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