Answer:
1.23 g
Explanation:
A chemist adds 1.55 L of a 0.00582 M calcium sulfate solution to a reaction flask. Calculate the mass in grams of calcium sulfate the chemist has added to the flask.
Step 1: Given data
Volume of the solution (V): 1.55 LMolar concentration of the solution (C): 0.00582 M (0.00582 mol/L)Step 2: Calculate the moles (n) of calcium sulfate added
We will use the following expression.
n = C × V
n = 0.00582 mol/L × 1.55 L
n = 0.00902 mol
Step 3: Calculate the mass corresponding to 0.00902 moles of calcium sulfate
The molar mass of calcium sulfate is 136.14 g/mol.
0.00902 mol × 136.14 g/mol = 1.23 g
if you are provided with 34.5 g iron (iii) chloride and 25.3 g sodium carbonate what is the limiting ractant
Answer:
Sodium carbonate, Na₂CO₃ is the limiting
Explanation:
We'll begin by writing the balanced equation for the reaction between iron (iii) chloride, FeCl₃ and sodium carbonate, Na₂CO₃.
This is illustrated below:
2FeCl₃ + 3Na₂CO₃ → Fe₂(CO₃)₃ + 6NaCl
Next, we shall determine the masses of FeCl₃ and Na₂CO₃ that reacted from the balanced equation. This can be obtained as follow:
Molar mass of FeCl₃ = 56 + (3×35.5)
= 56 + 106.5
= 162.5 g/mol
Mass of FeCl₃ from the balanced equation = 2 × 162.5 = 325 g
Molar mass of Na₂CO₃ = (2×23) + 12 + (3×16)
= 46 + 12 + 48
= 106 g/mol
Mass of Na₂CO₃ from the balanced equation = 3 × 106 = 318 g
SUMMARY:
From the balanced equation above,
325 g of FeCl₃ reacted with 318 g of Na₂CO₃.
Finally, we shall determine the limiting reactant. This can be obtained as follow:
From the balanced equation above,
325 g of FeCl₃ reacted with 318 g of Na₂CO₃.
Therefore, 34.5 g of FeCl₃ will react with = (34.5 × 318)/325 = 33.76 g of Na₂CO₃.
From the calculations made above, we can see clearly that it will take a higher amount (i.e 33.76 g) of Na₂CO₃ than what was given (i.e 25.3 g) to react completely with 34.5 g of FeCl₃. Therefore, Na₂CO₃ is the limiting reactant and FeCl₃ is the excess reactant.