Cal is titrating 57.5 mL of 0.363 M HBr with 0.315 M Ba(OH)2. How many mL of Ba(OH)2 does Cal need to add to reach the equivalence point?

33.1mL of 0.315M Ba(OH) are required to reach the equivalence point.

Explanation:

The reaction of HBr with Ba(OH)₂ is:

2HBr + Ba(OH)₂ →  BaBr₂ + 2H₂O

That means to find the theoretical volume Cal need to add we need to find moles of HBr in solution, convert these to moles of Ba(OH)₂ and with its molar concentration find the volume, thus:

Moles HBr:

57.5mL = 0.0575L * (0.363mol/L) = 0.0209 moles HBr

Moles Ba(OH)₂:

0.0209 moles HBr * (1mol Ba(OH)₂ / 2 mol HBr) = 0.0104 moles Ba(OH)₂

Volume:

0.0104 moles Ba(OH)₂ * (1L / 0.315mol) = 0.0331L =

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 !

a pan containing water is placed in a positio n where it is exposed to a 10 mile per hour wind. the pan contains a uniform depth of 1.27 cm of water, is quite wide, and has a length of 2.44 m in the wind direction. the water in the pan is at a constant temperature of 15.5 °c.