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GREYUIT [131]
3 years ago
8

A 2.567-g sample of a monoprotic acid was dissolved in water. It took 15.24 mL of a 0.1578 M NaOH solution to neutralize the aci

d. Calculate the molar mass of the acid.
Chemistry
1 answer:
olchik [2.2K]3 years ago
8 0

Answer:

The molar mass of the acid is 1067.42 g/mol  

Explanation:

A monotropic acid can donate one proton only in a acd base reaction thus

Monotropic acid + NaOH → NaSalt + H₂O

From the above reaction, one mole of NaOH reacts with one mole of monotropc acid

however there are 0.1578 M NaOH from 15.24 mL hence we have

Number of moles of NaOH = 0.1578 M/L × (15.24/1000) L = 0.002404872 M of NaOH

The number of moles of the monotropic acid = number of moles of NaOH = 0.0024 M. However Number of moles = mass/(molar mass)

Therefore molar mass = mass/number of moles = 2.567(2.4×10⁻³) = 1067.42 g/mol  

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Answer:

True statment

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Explanation:

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Hence

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Zigmanuir [339]

Answer:

0.453 moles

Explanation:

The balanced equation for the reaction is:

2Fe(s) + 3O2(g)  ==>  2Fe2O3

From the equation,  mass of O2 involved = 16 x 2 x 3 = 96g

                                 mass of Fe2O3 involved = [(2x26) + 3 x 16] x 2

                                                                            = 100g

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                                  32.2g of O2 Will produce   100x32.2/96

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Converting it to mole using   number of mole = mass/molar mass

but molar mass of Fe2O3 = 26 + (16 X 3)

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Therefore number of mole of 33.54g of Fe2O3 = 33.54/74

                                                                           = 0.453 moles

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