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Anastaziya [24]
2 years ago
15

An unknown diprotic acid (H2A) requires 44.391 mL of 0.111 M NaOH to completely neutralize a 0.58 g sample. Calculate the approx

imate molar mass of the acid.
Chemistry
1 answer:
Anna [14]2 years ago
5 0

Answer:

M=235.42g/mol

Explanation:

Hello!

In this case, given this is an acid-base neutralization and we are considering a diprotic acid, we can write the following mole-mole relationship:

2n_{acid}=n_{base}

It means that the moles of acid can be computed given the volume and concentration of NaOH:

n_{acid}=\frac{M_{base}V_{base}}{2} =\frac{0.044391L*0.111mol/L}{2} \\\\n_{acid}=2.46x10^{-4}mol

It means that the approximate molar mass of the acid is:

M=\frac{m_{acid}}{n_{acid}} \\\\M=\frac{m_{acid}}{n_{acid}} =\frac{0.58g}{2.46x10^{-3}mol}\\\\M=235.42g/mol

Best regards!

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Answer:

The answer to your question is below

Explanation:

Consider the reaction: P4 + 6Cl2 = 4PCl3.

a. How many grams of Cl2 are needed to react with 20.00 g of P4? ___68.7 g___________

                                P4      +      6Cl2      =      4PCl3

                          4(31) ---------- 12(35.5)

                         20     ----------    x

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                            P4      +      6Cl2      =      4PCl3

                       124g             426 g               4(31 + 3(35.5)) = 550g

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I will use P4 to find the limiting reactant

                 

                     x = (15 x 426) / 124 = 51.5   The limiting reactant is Chlorine

                                                                  because we need 51.5 g and we only have 22g

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