How many milliliters of 0.200M NaOHsolution are needed to react with 44.0 mL of a 0.300 M NiCl2 solution?

NiCl2 + 2NaOH ==> Ni(OH)2 + 2NaCl

mols NiCl2 = M x L = ?
Using the coefficients in the balanced equation, convert mols NiCl2 to mols NaOH.
Then M NaOH = mols NaOH/L NaOH. You know M and molsk solve for L NaOH and convert to mL.

To find the number of milliliters of 0.200M NaOH solution needed to react with 44.0 mL of a 0.300M NiCl2 solution, we can use the concept of stoichiometry and the balanced chemical equation for the reaction.

Let's assume that the balanced chemical equation is:

2NaOH + NiCl2 → Ni(OH)2 + 2NaCl

From the balanced equation, we can see that 2 moles of NaOH reacts with 1 mole of NiCl2.

First, let's calculate the number of moles of NiCl2 in 44.0 mL of a 0.300M NiCl2 solution.

Moles of NiCl2 = volume (in liters) × concentration
Moles of NiCl2 = (44.0 mL ÷ 1000 mL/L) × 0.300 mol/L
Moles of NiCl2 = 0.0132 mol

According to the stoichiometry of the balanced equation, 2 moles of NaOH are needed to react with 1 mole of NiCl2.

So, the number of moles of NaOH needed can be calculated using the mole ratio:

Moles of NaOH = (moles of NiCl2) × (2 moles of NaOH / 1 mole of NiCl2)
Moles of NaOH = 0.0132 mol × (2/1)
Moles of NaOH = 0.0264 mol

Now, let's calculate the volume of 0.200M NaOH solution needed to contain 0.0264 moles:

Volume of NaOH solution (in liters) = moles of NaOH / concentration
Volume of NaOH solution = 0.0264 mol / 0.200 mol/L
Volume of NaOH solution = 0.132 L

To convert the volume from liters to milliliters, multiply by 1000:

Volume of NaOH solution = 0.132 L × 1000 mL/L
Volume of NaOH solution = 132 mL

Therefore, you would need 132 milliliters of 0.200M NaOH solution to react with 44.0 mL of a 0.300M NiCl2 solution.