Answer:
Mass of excess reactant left = 48 g
Explanation:
Given data:
Moles of carbon = 3.00 mol
Moles of oxygen = 3.00 mol
Mass of excess reactant left = ?
Solution:
Chemical equation:
2C + O₂ → 2CO
Now we will compare the moles of C and O₂ with CO to determine the excess reactant.
C : CO
2 : 2
3 : 3
O₂ : CO
1 : 2
3 : 2/1×3 = 6
Number of moles of CO produced by C are less thus it will act as limiting reactant. Oxygen is in excess.
Now we will compare the moles of oxygen and carbon to calculate the moles of oxygen used.
C : O₂
2 : 1
3 : 1/2×3 = 1.5 mol
Thus 1.5 moles of oxygen are used.
Number of moles of oxygen left = 3 - 1.5 = 1.5 mol
Mass of oxygen left:
Mass = number of moles × molar mass
Mass = 1.5 mol × 32 g/mol
Mass = 48 g
A group goes vertically and a period goes horizontally.
(a group goes down and a period goes across)
If my friend was to say that I would say that the prediction are scientifically drawn. They look at the current state of the atmosphere. They also make educated guesses regarding previous weather patterns in that given area.
Answer: 36.53g
Explanation:
First we need to find the amount of NaCl that dissolves in 1L of the solution that produced 5M of NaCl
Molarity = 5M
MM of NaCl = 58.45
Molarity = Mass conc (g/L) / MM
Mass conc. (g/L) of NaCl = Molarity x MM
= 5 x 58.45 = 292.25g
Next, we need to find the amount that will dissolve in 125mL(i.e 0.125L)
From the calculations above,
292.25g of NaCl dissolved in 1L
Therefore Xg of NaCl will dissolve in 0.125L of the solution i.e
Xg of NaCl = 292.25 x 0.125 = 36.53g.
Therefore 36.53g of NaCl will dissolve in 125mL of the solution