Calculate the volume of a 0.225M solution of KOH required to react with 0.215g of acetic acid.

V = 0.018 L

Explanation:

Given data:

Mass of acetic acid = 0.215 g

Molarity of KOH = 0.225 M

Volume of KOH required = ?

Solution:

Chemical equation:

CH₃COOH + KOH      →    CH₃COOK + H₂O

Number of moles of acetic acid:

Number of moles = mass/molar mass

Number of moles = 0.215 g / 60.0 g/mol

Number of moles = 0.004 mol

now we will compare the moles of acetic acid and KOH.

                          CH₃COOH       :       KOH

                                  1                :          1

                              0.004           :       0.004

Volume of KOH required:

Molarity = number of moles / volume

0.255 M = 0.004 mol / V

V = 0.004 mol / 0.225 mol/L

V = 0.018 L

the answer is d   for sure : )

2fe + 3h2so4 -> fe2(so4)3 + 3h2

it is necessary to balance equations in order to get the proper amount of reactants as well as products in order not to make a miscalculation.

you cannot change the subscripts as that would change the substance itself. for instance, diatomic elements such as h2 only have the subscript two when a gas. another example would be so4, if you were to change the subscript, the element would no longer be sulfate.

we change the coefficients instead as that only changes the number of molecules rather than the substance itself.

hope this !