Answer:
C C2H5OH(l) + 3O2(g) → 2CO2 + 3H2O
Explanation:
La ley de conservación de la materia implica que en una reacción química los átomos en los reactivos son iguales a la cantidad de átomos en los productos:
A C2H5OH(l) + O2(g) → CO2 + H2O
En los reactivos hay dos átomos de carbono pero en los productos solo 1. De esta manera, no cumple la ley de conservación.
B C2H5OH(l) + O2(g) → 2CO2 + H2O
En los reactivos hay 6 átomos de hidrógeno pero en los productos solo 2. No cumple la ley de conservación.
C C2H5OH(l) + 3O2(g) → 2CO2 + 3H2O
En los reactivos y productos hay: 2 átomos de carbono, 6 átomos de hidrógeno y 7 átomos de oxígeno. Cumple la ley de conservación.
D C2H5OH(l) + O2(g) → 2CO2 + 3H2O
En los reactivos hay 3 átomos de oxígeno y en los productos 7. No cumple la ley de conservación.
E 2C2H5OH(l) + 3O2(g) → 2CO2 + 3H2O
En los reactivos hay 4 átomos de carbono y en los productos solo dos. No cumple la ley de conservación.
Answer:
None of the options are correct. The correct answer is:
56.67g
Explanation:
N2 + 3H2 —> 2NH3
Molar Mass of NH3 = 14 + (3x1) = 14 + 3 = 17g/mol
Mass of NH3 from the balanced equation = 2 x 17 = 34g
Molar Mass of H2 = 2x1 = 2g/mol
Mass of H2 from the balanced equation = 3 x 2 = 6g
From the equation,
6g of H2 produced 34g of NH3.
Therefore, 10g of H2 will produce = (10 x 34)/6 = 56.67g of NH3
Therefore, 56.67g of NH3 are produced
Answer: jjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjj
Explanation:
The mole fraction of KBr in the solution is 0.0001
<h3>How to determine the mole of water</h3>
We'll begin by calculating the mass of the water. This can be obtained as follow:
- Volume of water = 0.4 L = 0.4 × 1000 = 400 mL
- Density of water = 1 g/mL
- Mass of water =?
Density = mass / volume
1 = Mass of water / 400
Croiss multiply
Mass of water = 1 × 400
Mass of water = 400 g
Finally, we shall determine the mole of the water
- Mass of water = 400 g
- Molar mass of water = 18.02 g/mol
- Mole of water = ?
Mole = mass / molar mass
Mole of water = 400 / 18.02
Mole of water = 22.2 moles
<h3>How to de terminethe mole of KBr</h3>
- Mass of KBr = 0.3 g
- Molar mass of KBr = 119 g/mol
- Mole of KBr = ?
Mole = mass / molar mass
Mole of KBr = 0.3 / 119
Mole of KBr = 0.0025 mole
<h3>How to determine the mole fraction of KBr</h3>
- Mole of KBr = 0.0025 mole
- Mole of water = 22.2 moles
- Total mole = 0.0025 + 22.2 = 22.2025 moles
- Mole fraction of KBr =?
Mole fraction = mole / total mole
Mole fraction of KBr = 0.0025 / 22.2025
Mole fraction of KBr = 0.0001
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