1) What mass of Na2CO3 is required to make 50cc of its seminormal solution?
1 answer:
Answer:
Explanation:
Hello,
In this case, by considering the given seminormal solution, we infer it is a 0.5-N solution which means that we can obtain the equivalent grams as shown below for the 55 cc (0.055 L) volume:
Next, since sodium carbonate has two sodium ions with a +1 oxidation state each, we can obtain the moles:
Finally, the mass is computed by using its molar mass (106 g/mol)
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Answer:
2.52L
Explanation:
Given parameters:
T₁ = 400K
V₁ = 4L
T₂ = 252K
unknown
V₂ = ?
Solution:
To solve this problem, we are going to apply charle's law. The law states that the volume of a fixed mass of gas is directly proportional to temperature provided pressure is constant.
Mathematically,
Substitute and solve for V₂
V₂ = 2.52L
Answer:
Explanation:
Given parameters:
Mass of aluminium oxide = 3.87g
Mass of water = 5.67g
Unknown:
Limiting reactant = ?
Solution:
The limiting reactant is the reactant in short supply in a chemical reaction. We need to first write the chemical equation and convert the masses given to the number of moles.
Using the number of moles, we can ascertain the limiting reactants;
Al₂O₃ + 3H₂O → 2Al(OH)₃
Number of moles;
Number of moles =
molar mass of Al₂O₃ = (2x27) + 3(16) = 102g/mole
number of moles = = 0.04mole
molar mass of H₂O = 2(1) + 16 = 18g/mole
number of moles = = 0.32mole
From the reaction equation;
1 mole of Al₂O₃ reacted with 3 moles of H₂O
0.04 mole of Al₂O₃ will react with 3 x 0.04 mole = 0.12 mole of H₂O
But we were given 0.32 mole of H₂O and this is in excess of amount required.
This shows that Al₂O₃ is the limiting reactant