A 0.100 g sample of phosphoric acid (H3PO4, MW 99.0 g/mol), a tri-protic acid, requires 35.0 mL of a NaOH solution for neutralization. What is the molarity of the NaOH solution

Answers

Answer 1

Answer:

0.0866 M

Explanation:

answer key

Answer 2

The molarity of the NaOH solution has been 0.0865 M.

The neutralization reaction of the Phosphoric acid and NaOH results in the formation of sodium phosphate and water.

The balanced chemical equation can be given as;

[tex]\rm H_3PO_4\;+\;3\;NaOH\;\rightarrow\;Na_3PO_4\;+\;3\;H_2O[/tex]

Thus for the neutralization of 1 mole of Phosphoric acid, 3 moles of NaOH has been used.

The moles of phosphoric acid in 0.100 grams sample:

Moles = [tex]\rm \dfrac{weight}{molecular\;weight}[/tex]

Moles of phosphoric acid = [tex]\rm \dfrac{0.100}{99}[/tex]

Moles of phosphoric acid = 0.001 mol

For the neutralization of 0.001 moles of NaOH:

1-mole Phosphoric acid = 3 moles NaOH

0.001-mole Phosphoric acid =0.003 moles of NaOH.

The molarity of the NaOH solution:

Molarity = [tex]\rm \dfrac{moles}{Volume\;(L)}[/tex]

Molarity of NaOH = [tex]\rm \dfrac{0.003}{0.035}[/tex]

Molarity of NaOH = 0.0865 M

The molarity of the NaOH solution has been 0.0865 M.

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Take 56 mL and subtract the original volume of 50 mL and that will give you the volume of the rock, which is 6mL.

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Calculate the normality of a solution made by diluting 12 g of H3PO4 with water to produce 250.0 mL of solution.

Answers

The  normality of a solution made by diluting 12 g of H3PO4 with water to produce 250.0 mL of solution is 1.47 eq/L.

The normality of a solution is;

N = Mass of solute in grams/volume of solution in litres × Equivalent weight

Mass of solute =  12 g

volume of solution =  250.0 mL or 0.25 L

Equivalent weight = Molar mass of H3PO4/n

n = Number of hydrogen ions in H3PO4

Equivalent weight = 98 g/mol/3 = 32.7

Substituting values;

N = 12/0.25 ×  32.7

N = 1.47 eq/L

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calculate the mass of butane needed to produce 64.1 g of carbon dioxide to three significant figures and appropriate units

Answers

Answer:

21.2 gm

Explanation:

calculate the mass of butane needed to produce 64.1 g of carbon dioxide to three significant figures and appropriate units

butane is the hydrocarbon C4H10  

in combustion, we react hydrocarbons with O2 to form CO2 and H2O

so

C4H10  + O2---------------->  CO2 + H2O

BALANCE

2C4H10 + 1302--------> 8CO2 + 10 H2O

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64.1 gm of CO2 is

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SO

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Answers

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Hope this helps!

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Answer:

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Answers

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Answers

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Answer:

Hey mate.....

Explanation:

This is ur answer.....

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