how many milliliters of a 2.61 m h2so4 solution are needed to neutralize 71.50 ml of a 0.855 m koh solution
Apparently you don't know where the caps key is. You should find it because 2.61 m and 2.61M don't meant the same thing. I will assume that is 2.61 M and 0.855 M.
2KOH + H2SO4 ==> K2SO4 + 2H2O
mols KOH = M x L = ?
mols H2SO4 = 1/2 mols KOH. Look at the coefficients in the balanced equation.
Then M H2SO4 = mols H2SO4/L H2SO4. You know mols and M, solve for L and convert to mL.
To find out how many milliliters of a 2.61 M H2SO4 solution are needed to neutralize 71.50 mL of a 0.855 M KOH solution, we can use the concept of stoichiometry.
First, let's determine the balanced chemical equation for the neutralization reaction between sulfuric acid (H2SO4) and potassium hydroxide (KOH):
H2SO4 + 2 KOH → K2SO4 + 2 H2O
From the balanced equation, we can see that one mole of H2SO4 reacts with 2 moles of KOH.
To calculate the number of moles of KOH in the given 71.50 mL of 0.855 M KOH solution, we use the formula:
moles = concentration (M) × volume (L)
Converting the volume from milliliters to liters:
volume (L) = 71.50 mL × (1 L / 1000 mL) = 0.07150 L
Now, we can calculate the moles of KOH:
moles of KOH = 0.855 M × 0.07150 L = 0.06107 moles
Since the stoichiometric ratio between H2SO4 and KOH is 1:2, we need twice the number of moles of H2SO4 to neutralize the given amount of KOH.
Therefore, the number of moles of H2SO4 needed is:
moles of H2SO4 = 2 × 0.06107 moles = 0.1221 moles
Now, to determine the volume (in milliliters) of the 2.61 M H2SO4 solution needed to provide 0.1221 moles, we rearrange the moles formula:
volume (L) = moles / concentration (M)
Converting to milliliters:
volume (mL) = 0.1221 moles / 2.61 M × (1000 mL / 1 L) = 46.8 mL
Therefore, approximately 46.8 milliliters of the 2.61 M H2SO4 solution are needed to neutralize 71.50 milliliters of the 0.855 M KOH solution.