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Taking into account the reaction stoichiometry, 1.729 grams of NaCl is formed.
<h3>Reaction stoichiometry</h3>In first place, the balanced reaction is:
CuCl₂ + 2 NaNO₃ → Cu(NO₃)₂ + 2 NaCl
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
- CuCl₂: 1 mole
- NaNO₃: 2 moles
- Cu(NO₃)₂ : 1 mole
- NaCl: 2 moles
The molar mass of the compounds is:
- CuCl₂: 134.44 g/mole
- NaNO₃: 85 g/mole
- Cu(NO₃)₂ : 187.54 g/mole
- NaCl: 58.45 g/mole
Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:
- CuCl₂: 1 mole ×134.44 g/mole= 134.44 grams
- NaNO₃: 2 moles ×85 g/mole= 170 grams
- Cu(NO₃)₂ : 1 mole ×187.54 g/mole= 187.54 grams
- NaCl: 2 moles ×58.45 g/mole= 116.9 grams
The following rule of three can be applied: If by stoichiometry of the reaction 134.44 grams of CuCl₂ form 116.9 grams of NaCl, 2 grams of CuCl₂ form how much mass of NaCl?
<u><em>mass of NaCl= 1.739 grams</em></u>
Finally, 1.729 grams of NaCl is formed.
Learn more about the reaction stoichiometry:
Answer:
The relative rates of diffusion for methane and oxygen is 1.4142.
Methane gas will be able to travel 1.4142 meter in the same conditions.
Explanation:
To calculate the rate of diffusion of gas, we use Graham's Law.
This law states that the rate of effusion or diffusion of gas is inversely proportional to the square root of the molar mass of the gas. Mathematically written as:
We are given:
Molar mass of methane gas, m = 16 g/mol
Molar mass of oxygen gas,m' = 32 g/mol
By taking their ratio, we get:
The relative rates of diffusion for methane and oxygen is 1.4142.
If oxygen gas travels 1 meters in time t.
Rate of diffusion of oxygen =
If methane gas travels travels in y meters in time t.
Rate of diffusion of methane=
y = 1.4142 m
Methane gas will be able to travel 1.4142 meter in the same conditions.