How many liters of 0.300 M aluminum (III) hydroxide do you need to titrate 0.300 L of a 0.300 M diprotic acid to the equivalence point

Question

How many liters of 0.300 M aluminum (III) hydroxide do you need to titrate 0.300 L of a 0.300 M diprotic acid to the equivalence point

in progress 0
RI SƠ 5 years 2021-08-22T23:46:32+00:00 1 Answers 17 views 0

Answers ( )

  1. haidang
    0
    2021-08-22T23:47:37+00:00
    Reply

    Answer:

    0.200L of 0.300 Al(OH)₃ are necessaries

    Explanation:

    A diprotic acid, H₂X, reacts with aluminium (III) hydroxide, Al(OH)₃ as follows:

    3 H₂X + 2 Al(OH)₃ →  6H₂O + Al₂X₃

    To solve this question we must find the moles of the H₂X, using the balanced reaction we can find the moles of Al(OH)₃ and its volumen knowing its molar concentration is 0.300M:

    Moles H₂X:

    0.300L * (0.300mol / L) = 0.0900moles H₂X

    Moles Al(OH)₃:

    0.0900moles H₂X * (2mol Al(OH)₃ / 3mol H₂X) = 0.0600 moles Al(OH)₃

    Volume 0.300M Al(OH)₃:

    0.0600 moles Al(OH)₃ * (1L / 0.300moles) =

    0.200L of 0.300 Al(OH)₃ are necessaries

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )