A piston starts out with a volume of 3.000 L at 250.0 K and 800.0 torr. It is heated to 500.0 K and has a new volume of 4.000 L.
1 answer:
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Answer:
71.5g
Explanation:
The reaction equation is given as:
C + O₂ → CO₂
Mass of C = 42g
Mass of O₂ = 52g
Unknown:
Mass of CO₂ produced = ?
Solution
Now to solve this problem, we have to find limiting reactant which is the one given in short supply in this reaction.
The extent of the reaction is controlled by this reactant.
Find the number of moles of the given species;
Number of moles =
Number of moles of C = = 3.5mol
Number of moles of O₂ = = 1.63mol
Now;
From the balanced reaction equation;
1 mole of C reacted with 1 mole of O₂
We see that C is in excess and O₂ is the limiting reactant.
1 mole of O₂ will produce 1 mole of CO₂
So; 1.63mole of O₂ will produce 1.63 mole of CO₂
Mass of CO₂ = number of moles x molar mass
Molar mass of CO₂ = 44g/mol
Mass of CO₂ = 1.63 x 44 = 71.5g
A saturated solution is one in which no more solute is able to dissolve in a given solvent at a particular temperature. Some amount of the solute is left undissolved in the solution.
Unsaturated solution has solute in lower proportions than required to form a saturated solution.
Supersaturated solution has solute in amounts greater than a saturated solution.
We can take the help of solubility curve in order to find out the amount of a salt required to prepare a saturated solution of that salt at a particular temperature.
The solubility of KI at 10 is 136 g/ 100 mL water
The solubility of at
is 21 g/100 mL water.
The solubility of at
is 80 g/100 mL water.
The solubility of NaCl at is 38 g/ 100 mL water.
So the correct answer will be KI, as it would need 136 g KI / 100 mL water to form a saturated solution at .So, if we have 80g KI/ 100mL water it would be an unsaturated solution.
Answer:
599.26 grams of potassium sulfate will be produced.
Explanation:
Moles of chromium (III) sulfate =
According to reaction, 1 mole of chromium (III) sulfate gives 3 moles of potassium sulfate.
Then 1.1480 moles of chromium (III) sulfate will give:
Mass of 3.4440 moles of potassium sulfate:
= 3.4440 mol × 174 g/mol = 599.26 g
599.26 grams of potassium sulfate will be produced.