How many milliliters of a 0.212 m hi solution are needed to reduce 20.5 ml of a 0.358 m kmno4 solution according to the following equation: 10hi + 2kmno4 + 3h2so4 → 5i2 + 2mnso4 + k2so4 + 8h2o

Volume = 0.17 L

Explanation:

For :

Molarity = 0.358 M

Volume = 20.5 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 20.5×10⁻³ L

Moles of :

According to the reaction shown below:-

2 moles of reacts with 10 moles of HI

Also,

1 mole of reacts with 5 moles of HI

So,

0.007339 mole of reacts with 5*0.007339 moles of HI

Moles of HI = 0.036695 moles

Volume = ?

Molarity = 0.212 M

So,

Volume = 0.17 L

n2(g) + 3h2-> 2nh3(g)   this is the balanced equation

note the mole ratio between n2, h2 and nh3.   it is 1 : 3 : 2   this will be important.

moles n2 present = 28.0 g n2 x 1 mole n2/28 g = 1 mole n2 present

moles h2 present = 25.0 g h2 x 1 mole h2/2 g = 12.5 moles h2 present

based on mole ratio, n2 is limiting in this situation because there is more than enough h2 but not enough n2.

moles nh3 that can be produced = 1 mole n2 x 2 moles nh3/mole n2 = 2 moles nh3 can be produced

grams of nh3 that can be produced = 2 moles nh3 x 17 g/mole = 34 grams of nh3 can be produced

note:   the key to this problem is recognizing that n2 is limiting, and therefore limits how much nh3 can be produced.

what is the question here. i think you messed up.