The choices can be found elsewhere and as follows:
a. mass-mass problems
<span>b. mass-volume problems </span>
<span>c. mass-particle problems </span>
<span>d. volume-volume problems
</span>
I believe the correct answer is option D. It is volume-volume problems that does not require the use of molar mass. <span> Here you are dealing with molarities and volumes to determine concentrations. Molar mass is not part of any calculations.</span>
Answer:
The answer to your question is 242 ml
Explanation:
Data
HI 0.211 M Volume = x
KMnO₄ 0.354 M Volume = 24 ml
Balanced Chemical reaction
12HI + 2KMnO₄ + 2H₂SO₄ → 6I₂ + Mn₂SO₄ + K₂SO₄ + 8H₂O
Process
1.- Calculate the moles of KMnO₄ 0.354 M in 24 ml
Molarity = moles / volume (L)
moles = Molarity x volume (L)
moles = 0.354 x 0.024
moles = 0.0085
2.- From the balanced chemical reaction we know that HI and KMnO₄ react in the proportion 12 to 2. Then,
12 moles of HI --------------- 2 moles of KMnO₄
x --------------- 0.0085 moles of KMnO₄
x = (0.0085 x 12)/2
x = 0.051 moles of HI
3.- Calculate the milliliters of HI 0.211 M
Molarity = moles/volume
Volume = moles/molarity
Volume = 0.051/0.211
Volume = 0.242 L or Volume = 242 ml
The answer is c Yep Allll day
