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musickatia [10]
3 years ago
13

An acid with molar mass 84.48 g/mol is titrated with 0.650 M KOH. What volume of KOH solution is needed to titrate 1.70 grams of

the acid to the equivalence point.
Chemistry
1 answer:
RSB [31]3 years ago
7 0

Answer:

V=0.0310L=3.10mL

Explanation:

Hello.

In this case, since the acid is monoprotic and the KOH has one hydroxyl ion only, we can see that at the equivalence point the moles of both of them are the same:

n_{acid}=n_{KOH}

Thus, since we are given 1.70 g of the acid, we compute the moles of acid that were titrated:

n_{acid}=1.70g*\frac{1mol}{84.48g}=0.0201mol

Which equal the moles of KOH. In such a way, since the molarity is defined as moles over liters (M=n/V), the liters are moles over molarity (V=n/M), thus, the resulting volume is:

V=\frac{0.0201mol}{0.650mol/L}\\\\V=0.0310L=3.10mL

Best regards!

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How many grams of oxygen can be prepared by the decomposition of 12 grams of mercury oxide<br><br>​
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Taking into account the reaction stoichiometry, 0.886 grams of O₂ can be prepared by the decomposition of 12 grams of mercury oxide.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction of the decomposition of mercury oxide is:

2 HgO  → 2 Hg + O₂

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

  • HgO: 2 moles  
  • Hg: 2 moles
  • O₂: 1 mole

The molar mass of the compounds is:

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  • Hg: 200.59 g/mole
  • O₂: 32 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

HgO: 2 moles ×216.59 g/mole= 433.18 grams

Hg: 2 moles ×200.59 g/mole= 401.18 grams

O₂: 1 mole ×32 g/mole= 32 grams

<h3>Mass of oxygen formed</h3>

The following rule of three can be applied: if by reaction stoichiometry 433.18 grams of HgO form 32 grams of O₂, 12 grams of HgO form how much mass of O₂?

mass of O_{2} =\frac{12 grams of HgOx 32 grams of O_{2}}{433.18 grams of HgO}

<u><em>mass of O₂= 0.886 grams</em></u>

Then, 0.886 grams of O₂ can be prepared by the decomposition of 12 grams of mercury oxide.

Learn more about the reaction stoichiometry:

brainly.com/question/24741074

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