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2 years ago

How many moles of HNO3 are present if 4.90×10−2 mol of Ba(OH)2 was needed to neutralize the acid solution?

Chemistry

1 answer:

Amiraneli [1.4K]2 years ago

3 0

Answer:

0.098 moles

Explanation:

Let y represent the number of moles present

1 mole of Ba(OH)₂ contains 2 moles of OH- ions.

Hence, 0.049 moles of Ba(OH)2 contains y moles of OH- ions.

To get the y moles, we then do cross multiplication

1 mole * y mole = 2 moles * 0.049 mole

y mole = 2 * 0.049 / 1

y mole = 0.098 moles of OH- ions.

1 mole of OH- can neutralize 1 mole of H+

Therefore, 0.098 moles of HNO₃ are present.

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3 years ago

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2 years ago

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Answer:

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3 years ago

18.2 mL of a 0.156 M solution of lead(II) nitrate are added to 26.2 mL of a 0.274 M solution of potassium iodide.

kow [346]

The mass of the Pbl2 : 1308.87

<h3>Further explanation</h3>

Given

18.2 mL of a 0.156 M Pb(NO3)2

26.2 mL of a 0.274 M KI

Reaction

Pb(NO3)2 (aq) + 2 KI (aq) - Pbl2 (s) + 2 KNO3

Required

the mass of the Pbl2

Solution

mol Pb(NO3)2 = 18.2 x 0.156 = 2.8392 mlmol

mol KI = 26.2 x 0.274 =7.1788 mol

Limiting reactant Pb(NO3)2(smaller ratio of mol : reaction coeffiecient)

mol Pbl2 based on limiting reactant (Pb(NO3)2)

From equation, mol ratio of Pb(NO3)2 : Pbl2 = 1 : 1, so mol Pbl2=mol Pb(NO3)2=2.8392

Mass Pb(NO3)2 :

$\tt mass=mol\times MW\\\\mass=2.8392\times 461\\\\mass=1308.87~grams$

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3 years ago

Carbon can form large number of organic compounds why

Vinvika [58]

Answer:

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2 years ago