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Citrus2011 [14]
3 years ago
9

The percent yield for the reaction in which 15.6g of aluminum hydroxide is reacted in excess hydrogen chloride gas is 92.5%. Wha

t is the actual yield of aluminum chlorine?
Chemistry
1 answer:
Misha Larkins [42]3 years ago
4 0

Answer:

24.6 g

Explanation:

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Examine the given reaction. NH4NO3(s) → NH4+(aq) + NO3–(aq) ΔH° = 25.45 kJ/mol ΔS° = 108.7 J/mol·K Which of the given is correct
kobusy [5.1K]

Answer:

B)−6,942 J /mol

Explanation:

At constant temperature and pressure, you cand define the change in Gibbs free energy, ΔG, as:

ΔG = ΔH - TΔS

Where ΔH is enthalpy, T absolute temperature and ΔS change in entropy.

Replacing (25°C = 273 + 25 = 298K; 25.45kJ/mol = 25450J/mol):

ΔG = ΔH - TΔS

ΔG = 25450J/mol - 298K×108.7J/molK

ΔG = -6942.6J/mol

Right solution is:

<h3>B)−6,942 J /mol</h3>

8 0
3 years ago
Air is compressed from an inlet condition of 100 kPa, 300 K to an exit pressure of 1000 kPa by an internally reversible compress
ElenaW [278]

Answer:

(a) W_{isoentropic}=8.125\frac{kJ}{mol}

(b) W_{polytropic}=7.579\frac{kJ}{mol}

(c) W_{isothermal}=5.743\frac{kJ}{mol}

Explanation:

Hello,

(a) In this case, since entropy remains unchanged, the constant k should be computed for air as an ideal gas by:

\frac{R}{Cp_{air}}=1-\frac{1}{k}  \\\\\frac{8.314}{29.11} =1-\frac{1}{k}\\

0.2856=1-\frac{1}{k}\\\\k=1.4

Next, we compute the final temperature:

T_2=T_1(\frac{p_2}{p_1} )^{1-1/k}=300K(\frac{1000kPa}{100kPa} )^{1-1/1.4}=579.21K

Thus, the work is computed by:

W_{isoentropic}=\frac{kR(T_2-T_1)}{k-1} =\frac{1.4*8.314\frac{J}{mol*K}(579.21K-300K)}{1.4-1}\\\\W_{isoentropic}=8.125\frac{kJ}{mol}

(b) In this case, since n is given, we compute the final temperature as well:

T_2=T_1(\frac{p_2}{p_1} )^{1-1/n}=300K(\frac{1000kPa}{100kPa} )^{1-1/1.3}=510.38K

And the isentropic work:

W_{polytropic}=\frac{nR(T_2-T_1)}{n-1} =\frac{1.3*8.314\frac{J}{mol*K}(510.38-300K)}{1.3-1}\\\\W_{polytropic}=7.579\frac{kJ}{mol}

(c) Finally, for isothermal, final temperature is not required as it could be computed as:

W_{isothermal}=RTln(\frac{p_2}{p_1} )=8.314\frac{J}{mol*K}*300K*ln(\frac{1000kPa}{100kPa} ) \\\\W_{isothermal}=5.743\frac{kJ}{mol}

Regards.

8 0
3 years ago
How is a metallic compound different from an ionic compound?
nata0808 [166]

Answer:

Ionic compounds are compounds whereas metals are elements. Ionic compounds are electrolytes whereas metals are conductors. Ionic compounds are brittle whereas metals are malleable and ductile. ... Most ionic compounds will dissolve in polar solvents like water whereas metals will either be insoluble or react with water.

8 0
3 years ago
Which phase of matter has a fixed volume but variable shape?
jek_recluse [69]
The answer is "Liquid".

Hope I could help you!
Have a great day ;)
8 0
3 years ago
Read 2 more answers
1) When 2.38g of magnesium is added to 25.0cm of 2.27 M hydrochloric acid, hydrogen gas is released.
Andrej [43]

Answer:

a. HCl.

b. 0.057 g.

c. 1.69 g.

d. 77 %.

Explanation:

Hello!

In this case, since the reaction between magnesium and hydrochloric acid is:

Mg+2HCl\rightarrow MgCl_2+H_2

Whereas there is 1:2 mole ratio between them.

a) Here, we can identify the limiting reactant as that yielded the fewest moles of hydrogen gas product via the 1:1 and 2:1 mole ratios:

n_{H_2}^{by\  HCl}=0.025L*2.27\frac{molHCl}{1L}*\frac{1molH_2}{2molHCl}  =0.0284molH_2\\\\n_{H_2}^{by\  Mg}=2.38gMg*\frac{1molMg}{24.3gMg}*\frac{1molH_2}{1molMg}=0.0979molH_2

Thus, since hydrochloric yields fewer moles of hydrogen than magnesium, we realize it is the limiting reactant.

b) Here, we use the molar mass of gaseous hydrogen (2.02 g/mol) to compute the mass:

m_{H_2}=0.0284molH_2*\frac{2.02gH_2}{1molH_2}=0.057gH_2

c) Here, we compute the mass of magnesium associated with the yielded 0.0248 moles of hydrogen:

m_{Mg}^{reacted}=0.0284molH_2*\frac{1molMg}{1molH_2}*\frac{24.3gMg}{1molMg}  =0.690gMg

Thus, the mass of excess magnesium turns out:

m_{Mg}^{excess}=2.38g-0.690g=1.69gMg

d) Finally, we compute the percent yield, considering 0.044 g is the actual yield and 0.057 g the theoretical yield:

Y=\frac{0.044g}{0.057g} *100\%\\\\Y=77\%

Best regards!

8 0
3 years ago
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