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Tems11 [23]
2 years ago
11

A 0.2688 g sample of a monoprotic acid neutralized 16.4 mL of 0.08133 M KOH solution. Calculate the molar mass of the acid

Chemistry
1 answer:
ivanzaharov [21]2 years ago
3 0

Answer:

202 g/mol

Explanation:

Let's consider the neutralization between a generic monoprotic acid and KOH.

HA + KOH → KA + H₂O

The moles of KOH that reacted are:

0.0164 L × 0.08133 mol/L = 1.33 × 10⁻³ mol

The molar ratio of HA to KOH is 1:1. Then, the moles of HA that reacted are 1.33 × 10⁻³ moles.

1.33 × 10⁻³ moles of HA have a mass of 0.2688 g. The molar mass of the acid is:

0.2688 g/1.33 × 10⁻³ mol = 202 g/mol

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0.779 cube centimetres

Explanation:

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8 0
3 years ago
Read 2 more answers
How many moles of H2O will be produced from 26.0 g of H2022<br> Stoichiome
lukranit [14]

Answer:

                       0.7644 moles of H₂O

Explanation:

                    The balance chemical equation is as follow;

                                      2 H₂O₂ → 2 H₂O + O₂

To solve this problem we will first calculate the moles of H₂O₂ as,

                           Moles  =  Mass / M/Mass

                           Moles  =  26.0 g / 34.01 g/mol

                           Moles  =  0.7644 mol

Secondly,

According to equation,

                      2 moles of H₂O₂ produces  =  2 moles of H₂O

Hence,

                0.7644 mol of H₂O₂ will produce  =  X moles of H₂O

Solving for X,

                        X  =  2 mol × 0.7644 mol / 2 mol

                       X =  0.7644 moles of H₂O

5 0
3 years ago
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