A sample of NO2 in a 1550 mL metal cylinder at 70.26 kPa has its temperature changed from 480oC to -237oC while its volume is si
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Answer:
a) CO₂: <em>21,9%; </em>CO: <em>7,0%; </em>CH₄: <em>18,2%; </em>H₂: <em>0,8%; </em>N₂: <em>52,1%</em>
b) 24,09 g/mol
Explanation:
a) It is posible to obtain the composition of the gas mixture in weight% using molecular mass of each compound, thus:
12% CO₂× = <em>528,1 g</em>
6% CO× = <em>168,1 g</em>
27,3% CH₄× = <em>438,2 g</em>
9,9% H₂× = <em>20,0 g</em>
44,8% N₂× = <em>1254,4 g</em>
The total mass of the gas mixture is:
528,1g + 168,1g + 438,2g + 20,0g + 1254,4g = <em>2408,8 g</em>
Thus composition of the gas mixture in weight% is:
CO₂: ×100 = <em>21,9%</em>
CO: ×100 = <em>7,0%</em>
CH₄: ×100 = <em>18,2%</em>
H₂: ×100 = <em>0,8%</em>
N₂: ×100 = <em>52,1%</em>
b) The average molecular weight of the gas mixture is determined with mole % composition, thus:
0,12×44,01g/mol + 0,06×28,01g/mol + 0,273×16,05g/mol + 0,099×2,02g/mol + 0,448×28g/mol = <em>24,09 g/mol</em>
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Answer:
a) No. of moles of hydrogen needed = 5.4 mol
b) Grams of ammonia produced = 27.2 g
Explanation:
a)
No. of moles of nitrogen = 1.80 mol
1 mole of nitrogen reacts with 3 moles of hydrogen
1.80 moles of nitrogen will react with
= 1.80 × 3 = 5.4 moles of hydrogen
b)
No. of moles of hydrogen = 2.4 mol
It is given that nitrogen is present in sufficient amount.
3 moles of hydrogen produce 2 moles of
2.4 moles of hydrogen will produce
=
Molar mass of ammonia = 17 g/mol
Mass in gram = No. of moles × Molar mass
Mass of ammonia in g = 1.6 × 17
= 27.2 g