Answer:
Explanation:
We will need a balanced equation with masses, moles, and molar masses, so let’s gather all the information in one place.
Mᵣ: 44.01
C₃H₈ + 5O₂ ⟶ 3CO₂ + 4H₂O
n/mol: 1.5
1. Calculate the moles of CO₂
The molar ratio is 3 mol CO₂:1 mol C₃H₈
2. Calculate the mass of CO₂.
Answer:
Oxygen is limiting reactant
Explanation:
2 H2 + O2 ======> 2 H2 O
from this equation (and periodic table) you can see that
4 gm of H combine with 32 gm O2
H / O = 4/32 = 1/8
32 /16 = 2/1 shows O is limiter
for 32 gm H you will need 256 gm O and you only have 16 gm
What do you mean is that a school question
Answer:
The correct answer is 1.33 x 10⁻⁵ M
Explanation:
The concentration of the stock solution is: C= 1.33 M
In the first dilution, the student added 1 ml of stock solution to 9 ml of water. The total volume of the solution is 1 ml + 9 ml = 10 ml. So, the first diluted concentration is:
C₁= 1.33 M x 1 ml/10 ml = 1.33 M x 1/10 = 0.133 M
The second dilution is performed on C₁. The student added 1 ml of 0.133 M solution to 9 ml of water. Again, the total volume is 1 ml + 9 ml = 10 ml. The second diluted concentration is:
C₂= 0.133 M x 1 ml/10 ml = 0.133 M x 1/10= 0.0133 M
Since the student repeated the same dilution process 3 times more (for a total of 5 times), we have to multiply 5 times the initial concentration by the factor 1/10:
Final concentration = initial concentration x 1/10 x 1/10 x 1/10 x 1/10 x 1/10
= initial concentration x (1/10)⁵
= 1.33 M x 1 x 10⁻⁵
= 1.33 x 10⁻⁵ M
