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AnnyKZ [126]
3 years ago
9

A 0.1873 g sample of a pure, solid acid, H2X (a diprotic acid) was dissolved in water and titrated with 0.1052 M NaOH solution.

If the molar mass of H2X is 85.00 g/mol, calculate the volume of NaOH solution needed in the titration.
Chemistry
1 answer:
puteri [66]3 years ago
6 0

Answer:

We need 41.8 mL of NaOH

Explanation:

<u>Step 1:</u> Data given

Mass of H2X = 0.1873 grams

Molarity of NaOH solution = 0.1052 M

Molar mass of H2X = 85.00 g/mol

<u>Step 2</u>: The balanced equation

H2X (aq) +2 NaOH (aq) → Na2X (aq) + 2H2O(l)

<u>Step 3:</u> Calculate moles of H2X

Moles H2X = mass H2X / Molar mass H2X

Moles H2X = 0.1873 grams / 85.00 g/mol

Moles H2X = 0.0022 moles

<u>Step 4:</u> Calculate moles of NaOH

For 1 mol H2X we need 2 moles NaOH to produce 1 mole of Na2X and 2 moles of H2O

For 0.0022 moles of H2X we need 0.0044 moles of NaOH

<u />

<u>Step 5</u>: Calculate volume of NaOH

Volume of NaOH = moles of NaOH / molarity of NaOH

Volume of NaOH = 0.0044 moles / 0.1052 M

Volume NaOH =  0.0418 L = 41.8 mL

We need 41.8 mL of NaOH

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Black_prince [1.1K]
The mass  of water  produced  when  4.86 g  of  butane(C4H10)   react  with  excess oxygen  is  calculated  as below

calculate  the  moles of  C4H10 used = mass/molar mass

moles = 4.86g/58  g/mol =0.0838  moles
write a balanced equation   for  reaction

2 C4H10  + 13 O2 =  8 CO2  + 10 H2O
by use of mole   ratio between C4H10  to H2O  which is   2:10  the  moles  of
H20=  0.0838  x10/2 = 0.419  moles  of  H2O

mass = moles  x  molar mass

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6 0
3 years ago
Calculate how many grams of sodium azide (NaN3) are needed to inflate a 25.0 × 25.0 × 20.0 cm bag to a pressure of 1.35 atm at a
Luden [163]

Answer : The mass of NaN_3 at temperature 20^oC 28.47 g.

The mass of NaN_3 at temperature 10^oC 29.51 g.

Solution : Given,

Pressure of gas = 1.35 atm

Temperature of gas = 20^oC=273+20=293K     (0^oC=273K)

Volume of gas = 25\times 25\times 20cm=12500cm^3=12.5L   (1L=1000cm^3)

Molar mass of NaN_3 = 65 g/mole

Part 1 : First we have to calculate the moles of gas at temperature 20^oC. The gas produced in the given reaction is N_2.

Using ideal gas equation,

PV=nRT

where,

P = pressure of the gas

V = volume of the gas

T = temperature of the gas

n = number of moles of gas

R = Gas constant = 0.0821 Latm/moleK

Now put all the given values in this formula, we get

(1.35atm)\times (12.5L)=n\times (0.0821Latm/moleK)\times (293K)

By rearranging the terms, we get the value of 'n'

n=0.7015moles

The moles of N_2 = 0.7015 moles

The given balanced reaction is,

20NaN_3(s)+6SiO_2(s)+4KNO_3(s)\rightarrow 32N_2(g)+5Na_4SiO_4(s)+K_4SiO_4(s)

As, 32 moles of N_2 produced from 20 moles of NaN_3

So, 0.7015 moles of N_2 produced from \frac{20}{32}\times 0.7015=0.438 moles of NaN_3

Now we have to calculate the mass of NaN_3.

\text{ Mass of }NaN_3=\text{ Moles of }NaN_3\times \text{ Molar mass of }NaN_3

\text{ Mass of }NaN_3=(0.438moles)\times (65g/mole)=28.47g

Therefore, the mass of NaN_3 needed are 28.47 g.

Part 2 : We have to calculate the moles of gas at temperature 10^oC and same volume & pressure.

Using ideal gas equation,

PV=nRT

Now put all the given values in this formula, we get

(1.35atm)\times (12.5L)=n\times (0.0821Latm/moleK)\times (283K)

By rearranging the terms, we get the value of 'n'

n=0.726moles

The moles of N_2 = 0.726 moles

The given balanced reaction is,

20NaN_3(s)+6SiO_2(s)+4KNO_3(s)\rightarrow 32N_2(g)+5Na_4SiO_4(s)+K_4SiO_4(s)

As, 32 moles of N_2 produced from 20 moles of NaN_3

So, 0.726 moles of N_2 produced from \frac{20}{32}\times 0.726=0.454 moles of NaN_3

Now we have to calculate the mass of NaN_3.

\text{ Mass of }NaN_3=\text{ Moles of }NaN_3\times \text{ Molar mass of }NaN_3

\text{ Mass of }NaN_3=(0.454moles)\times (65g/mole)=29.51g

Therefore, the mass of NaN_3 needed are 29.51 g.

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Artemon [7]

Answer:

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grandymaker [24]
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frosja888 [35]

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6 0
4 years ago
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