Answer:
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Explanation:
<h3>#I'm not sure</h3><h3>#happy learning</h3>
Answer:
V = 43.95 L
Explanation:
Given data:
Mass of CH₄ decomposed = 15.63 g
Volume of H₂O produced at STP = ?
Solution:
Chemical equation:
CH₄ + 2O₂ → 2H₂O + CO₂
Number of moles of CH₄:
Number of moles = mass/molar mass
Number of moles = 15.63 g/ 16 g/mol
Number of moles = 0.98 mol
Now we will compare the moles of H₂O with CH₄.
CH₄ : H₂O
1 : 2
0.98 : 2×0.98 = 1.96 mol
Volume of hydrogen:
PV = nRT
1 atm × V = 1.96 mol × 0.0821 atm.L/mol.K × 273.15 K
V = 43.95atm.L / 1atm
V = 43.95 L
Given:
175 kilograms of Methane (CH4) to be synthesized into Hydrogen Cyanide (HCN)
The balanced chemical equation is shown below:
2 CH4<span> + 2 NH</span>3<span> + 3 O</span>2<span> → 2 HCN + 6 H</span>2<span>O
</span>
To calculate for the masses of ammonia and oxygen needed, our basis will be 175 kg CH4.
Molar mass:
CH4 = 16 kg/kmol
NH3 = 17 kg/kmol
O2 = 32 kg/kmol
mass of NH3 = 175 kg CH4 / 16 kg/kmol * (2/2) * 17 kg/kmol
mass of NH3 = 185.94 kg NH3 needed
mass of O2 = 175 kg CH4 / 16 kg/kmol * (3/2) * 32 kg/kmol
mass of O2 = 525 kg
mass of O = 525 kg / 32 kg/kmol * (1/2) * 16 kg/kmol
mass of O = 131.25 kg O
1.24973017189471 is probably the answer to your equation