Ask question

Ask question

All categories

3 years ago

10

A 1.555-g sample of baking soda decomposes with heat to produce 0.991 g Na2CO3. Refer to Example Exercise 14.l and show the calc

ulation for the theoretical yield of Na2CO3.
What is the percent yield of sodium carbonate, Na2CO3?
6. A 1473-g unknown mixture with baking soda is heated and has a mass loss of 0.325 g. Refer to Example Exercise 14.2 and show the calculation for the percentage NaHCOs in the mixture.

1 answer:

Sunny_sXe [5.5K]3 years ago

7 0

Answer:

a) 101%

b)59.7%

Explanation:

The equation for the thermal decomposition of baking soda is shown;

2NaHCO3 → Na2CO3 + H2O + CO2

Number of moles of baking soda= mass/molar mass= 1.555g/84.007 g/mol = 0.0185 moles

From the reaction equation;

2 moles of baking soda yields 1 mole of sodium carbonate

0.0185 moles of baking soda will yield = 0.0185 moles ×1 /2 = 9.25 ×10^-3 moles of sodium carbonate.

Therefore, mass of sodium carbonate= 9.25 ×10^-3 moles × 106gmol-1= 0.9805 g of sodium carbonate. This is the theoretical yield of sodium carbonate.

%yield = actual yield/theoretical yield ×100

% yield = 0.991/0.9805 ×100

%yield = 101%

Since ;

2NaHCO3 → Na2CO3 + H2O + CO2

And H2O + CO2 ---> H2CO3

Hence I can write, 2NaHCO3 → Na2CO3 + H2CO3

Molar mass of H2CO3= 62.03 gmol-1

Molar mass of baking soda= 84 gmol-1

Therefore, mass of baking soda=

0.325/62.03 × 2 × 84 = 0.88 g of NaHCO3

% of NaHCO3= 0.88/1.473 × 100 = 59.7%

You might be interested in

Which compound has the the highest melting point?

Kobotan [32]

The answer is
Al2(co3)3

5 0

3 years ago

Read 2 more answers

Please help on 44. /45./ 46./47,

Aleonysh [2.5K]

The only one that I can do without google is 47. Sorry that I can't answer the others. The answer to 47 is this: you know that the western side of the hill has the steepest slope because the ovals showing altitude are way closer together. The closer the circles/ovals are, the steeper the slope is.

Sorry if this doesn't help much, but I answered what I could without cheating.

Foxeslair

4 0

3 years ago

If you place 1.0 L of ethanol (C2H5OH) in a small laboratory that is 3.0 m long, 2.0 m wide, and 2.0 m high, will all the alcoho

ankoles [38]

If you place 1.0 L of ethanol (C2H5OH) in a small laboratory that is 3.0 m long, 2.0 m wide, and 2.0 m high, will all the alcohol evaporate? If some liquid remains, how much will there be? The vapor pressure of ethyl alcohol at 25 °C is 59 mm Hg, and the density of the liquid at this temperature is 0.785g/cm^3 .

will all the alcohol evaporate? or none at all?

Answer:

Yes, all the ethanol present in the laboratory will evaporate since the mole of ethanol present in vapor is greater. The volume of ethanol left will therefore be zero.

Explanation:

Given that:

The volume of alcohol which is placed in a small laboratory = 1.0 L

Vapor pressure of ethyl alcohol at 25 ° C = 59 mmHg

Converting 59 mmHg to atm ; since 1 atm = 760 mmHg;

Then, we have:

= $\frac{59}{760}atm$

= 0.078 atm

Temperature = 25 ° C

= ( 25 + 273 K)

= 298 K.

Density of the ethanol = 0.785 g/cm³

The volume of laboratory = l × b × h

= 3.0 m × 2.0 m × 2.5 m

= 15 m³

Converting the volume of laboratory to liter;

since 1 m³ = 100 L; Then, we have:

15 × 1000 = 15,000 L

Using ideal gas equation to determine the moles of ethanol in vapor phase; we have:

PV = nRT

Making n the subject of the formula; we have:

$n = \frac{PV}{RT}$

$n = \frac{0.078 * 15000}{0.082*290}$

n = 47. 88 mol of ethanol

Moles of ethanol in 1.0 L bottle can be calculated as follows:

Since numbers of moles = $\frac{mass}{molar mass}$

and mass = density × vollume

Then; we can say ;

number of moles = $\frac{density*volume }{molar mass of ethanol}$

number of moles = $\frac{0.785g/cm^3*1000cm^3}{46.07g/mol}$

number of moles = $\frac{&85}{46.07}$

number of moles = 17.039 mol

Thus , all the ethanol present in the laboratory will evaporate since the mole of ethanol present in vapor is greater. The volume of ethanol left will therefore be zero.

5 0

3 years ago

What physical property difference enabled you to Seperate salt from water?

elena55 [62]

<span>The two techniques for separating an insoluble solid from a liquid are filtration and centrifuging. Filtration relies on the fact that the solid particles are smaller than the filter paper pores which allow the tiny molecules of liquid to pass through. With solids which dissolve in a liquid solvent, evaporation is commonly used to recover the solid and distillation is used to recover the solvent.</span>

5 0

3 years ago

Read 2 more answers

Oxygen gas can be prepared by heating potassium chlorate according to the following equation:2KClO3(s)Arrow.gif2KCl(s) + 3O2(g)T

Eduardwww [97]

Answer:

Moles of potassium chlorate reacted = 0.2529 moles

The amount of oxygen gas collected will be 12.8675 g

Explanation:

(a)

We are given:

Vapor pressure of water = 17.5 mmHg

Total vapor pressure = 748 mmHg

Vapor pressure of Oxygen gas = Total vapor pressure - Vapor pressure of water = (748 - 17.5) mmHg = 730.5 mmHg

To calculate the amount of Oxygen gas collected, we use the equation given by ideal gas which follows:

$PV=nRT$

where,

P = pressure of the gas = 730.5 mmHg

V = Volume of the gas = 9.49 L

T = Temperature of the gas = $20^oC=[20+273]K=293K$

R = Gas constant = $62.3637\text{ L.mmHg }mol^{-1}K^{-1}$

n = number of moles of oxygen gas = ?

Putting values in above equation, we get:

$730.5mmHg\times 9.49L=n\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 293K\\\\n=\frac{730.5\times 9.49}{62.3637\times 293}=0.3794mol$

According to the reaction shown below as:-

$2KClO_3(s)\rightarrow 2KCl(s) +3O_2(g)$

3 moles of oxygen gas are produced when 2 moles of potassium chlorate undergoes reaction.

So,

0.3794 mol of oxygen gas are produced when $\frac{2}{3}\times 0.3794$ moles of potassium chlorate undergoes reaction.

<u>Moles of potassium chlorate reacted = 0.2529 moles</u>

(b)

We are given:

Vapor pressure of water = 17.5 mmHg

Total vapor pressure = 753 mmHg

Vapor pressure of Oxygen gas = Total vapor pressure - Vapor pressure of water = (753 - 17.5) mmHg = 735.5 mmHg

To calculate the amount of Oxygen gas collected, we use the equation given by ideal gas which follows:

$PV=nRT$

where,

P = pressure of the gas = 735.5 mmHg

V = Volume of the gas = 9.99 L

T = Temperature of the gas = $20^oC=[20+273]K=293K$

R = Gas constant = $62.3637\text{ L.mmHg }mol^{-1}K^{-1}$

n = number of moles of oxygen gas = ?

Putting values in above equation, we get:

$735.5mmHg\times 9.99L=n\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 293K\\\\n=\frac{735.5\times 9.99}{62.3637\times 293}=0.40211mol$

Moles of Oxygen gas = 0.40211 moles

Molar mass of Oxygen gas = 32 g/mol

Putting values in above equation, we get:

$0.037mol=\frac{\text{Mass of Oxygen gas}}{2g/mol}\\\\\text{Mass of Oxygen gas}=(0.40211mol\times 32g/mol)=12.8675g$

<u>Hence, the amount of oxygen gas collected will be 12.8675 g</u>

5 0

3 years ago