How many milliliters of 0.550 m hydriodic acid are needed to react with 25.00 ml of 0.217 m csoh?

Balanced equation is

HI + CsOH —> CsI + H2O
(Double replacement)

You can find the moles of CsOH by multiplying liters and molarity together. The formula is.

M = n / V
M is molarity, n is moles and is volume in liters

First change 25.00 mL to L by dividing 1000
• 0.025 L
Now multiply molarity and liters to get moles of CsOH

0.025 L x 0.217 M = 0.0054 moles CsOH

since the balanced equation is a one to one ratio, all compounds will be 0.0054 moles, in this case HI acid has 0.0054 moles.

Now you can find volume in liters by using the moles divided by the molarity of HI.

0.0054 moles HI / 0.550 M HI =
0.0098 L of HI

Now just change liters back to milliliters by multiplying by 1000

0.0098 x 1000 = 9.8 mL of HI needed

The problem here is that according to aufbau principle before 3d orbital 4s subshell comes.after filling 3p orbital with electrons you have 4 electrons left, you place two in 4s with opposite spin and the other two in 3d in any of its two orbitals of total five orbitals with the same spin.