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3 years ago

35g of HF was prepared by reacting 112g of CaF2 with an excess of H2SO4 calculate the percentage yield

Chemistry

1 answer:

Tom [10]3 years ago

7 0

Answer: 61%

The reaction equation should be

CaF2 + H2SO4 → 2HF + CaSO4

For every 1 molecule CaF2 used, there will be 2 molecules of HF formed. The molecular mass of CaF2 is 78/mol while the molecular mass of HF is 20g/mol. If the yield is 100%, the amount of HF formed by 112g CaF2 would be: 112g/(78g/mol) * 2 * (20g/mol)=57.43g

The percentage yield of the reaction would be: 35g/57.43g= 60.94%

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Using the thermodynamic information in the ALEKS Data tab, calculate the boiling point of titanium tetrachloride . Round your an

ddd [48]

Answer:

The boiling point is 308.27 K (35.27°C)

Explanation:

The chemical reaction for the boiling of titanium tetrachloride is shown below:

Ti $Cl_{4(l)}$ ⇒ Ti $Cl_{4(g)}$

ΔH° $_{f}$ (Ti $Cl_{4(l)}$ ) = -804.2 kJ/mol

ΔH° $_{f}$ (Ti $Cl_{4(g)}$ ) = -763.2 kJ/mol

Therefore,

ΔH° $_{f}$ = ΔH° $_{f}$ (Ti $Cl_{4(g)}$ ) - ΔH° $_{f}$ (Ti $Cl_{4(l)}$ ) = -763.2 - (-804.2) = 41 kJ/mol = 41000 J/mol

Similarly,

s°(Ti $Cl_{4(l)}$ ) = 221.9 J/(mol*K)

s°(Ti $Cl_{4(g)}$ ) = 354.9 J/(mol*K)

Therefore,

s° = s° (Ti $Cl_{4(g)}$ ) - s°(Ti $Cl_{4(l)}$ ) = 354.9 - 221.9 = 133 J/(mol*K)

Thus, T = ΔH° $_{f}$ /s° = [41000 J/mol]/[133 J/(mol*K)] = 308. 27 K or 35.27°C

Therefore, the boiling point of titanium tetrachloride is 308.27 K or 35.27°C.

5 0

3 years ago

Describe the differences between polyp and medusa

stealth61 [152]

One difference is that some animals are polyp and some are medusa.

The other difference is that some animals have medusa in their life or polp in their life cycle.

Hope these two differences helps :D

5 0

2 years ago

1. 17.0 grams of xenon hexafluoride is in a solid container. How many milliliters of that gas

BlackZzzverrR [31]

Answer: The volume of gas is 3020 ml

Explanation:

According to ideal gas equation:

$PV=nRT$

P = pressure of gas = 821.4 torr = 1.08 atm (760 torr = 1atm)

V = Volume of gas in L = ?

n = number of moles = $\frac{\text {given mass}}{\text {Molar mass}}=\frac{17.0g}{245.28g/mol}=0.069mol$

R = gas constant = $0.0821Latm/Kmol$

T =temperature = $302.7^0C=(302.7+273)K=575.7K$

$V=\frac{nRT}{P}$

$V=\frac{0.069mol\times 0.0821Latm/K mol\times 575.7K}{1.08atm}=3.02L=3020ml$

Thus volume of gas is 3020 ml

4 0

2 years ago

Be sure to answer all parts. The standard enthalpy of formation and the standard entropy of gaseous benzene are 82.93 kJ/mol and

skelet666 [1.2K]

Answer : The values of $\Delta H^o,\Delta S^o\text{ and }\Delta G^o$ are $33.89kJ,95.94J/K\text{ and }5.299kJ/mol$ respectively.

Explanation :

The given balanced chemical reaction is,

$C_6H_6(l)\rightarrow C_6H_6(g)$

First we have to calculate the enthalpy of reaction $(\Delta H^o)$ .

$\Delta H^o=H_f_{product}-H_f_{reactant}$

$\Delta H^o=[n_{C_6H_6(g)}\times \Delta H_f^0_{(C_6H_6(g))}]-[n_{C_6H_6(l)}\times \Delta H_f^0_{(C_6H_6(l))}]$

where,

$\Delta H^o$ = enthalpy of reaction = ?

n = number of moles

$\Delta H_f^0_{(C_6H_6(g))}$ = standard enthalpy of formation of gaseous benzene = 82.93 kJ/mol

$\Delta H_f^0_{(C_6H_6(l))}$ = standard enthalpy of formation of liquid benzene = 49.04 kJ/mol

Now put all the given values in this expression, we get:

$\Delta H^o=[1mole\times (82.93kJ/mol)]-[1mole\times (49.04J/mol)]$

$\Delta H^o=33.89kJ/mol=33890J/mol$

Now we have to calculate the entropy of reaction $(\Delta S^o)$ .

$\Delta S^o=S_f_{product}-S_f_{reactant}$

$\Delta S^o=[n_{C_6H_6(g)}\times \Delta S^0_{(C_6H_6(g))}]-[n_{C_6H_6(l)}\times \Delta S^0_{(C_6H_6(l))}]$

where,

$\Delta S^o$ = entropy of reaction = ?

n = number of moles

$\Delta S^0_{(C_6H_6(g))}$ = standard entropy of formation of gaseous benzene = 269.2 J/K.mol

$\Delta S^0_{(C_6H_6(l))}$ = standard entropy of formation of liquid benzene = 173.26 J/K.mol

Now put all the given values in this expression, we get:

$\Delta S^o=[1mole\times (269.2J/K.mol)]-[1mole\times (173.26J/K.mol)]$

$\Delta S^o=95.94J/K.mol$

Now we have to calculate the Gibbs free energy of reaction $(\Delta G^o)$ .

As we know that,

$\Delta G^o=\Delta H^o-T\Delta S^o$

At room temperature, the temperature is $25^oC\text{ or }298K$ .

$\Delta G^o=(33890J)-(298K\times 95.94J/K)$

$\Delta G^o=5299.88J/mol=5.299kJ/mol$

Therefore, the values of $\Delta H^o,\Delta S^o\text{ and }\Delta G^o$ are $33.89kJ,95.94J/K\text{ and }5.299kJ/mol$ respectively.

6 0

3 years ago

Read 2 more answers

PLS HELP!!

Gnesinka [82]

Answer:

from the 1st equation:

4NH3 4NO

4 *(68) 4*30

1216 X mass of NO = 536.5 g

from the 2nd Equation

2NO 2NO2

2*30 2* 46

536.5 x mass of NO2 = 822.6 grams

from the 3rd Equation

3NO2 2HNO3

3*(46) 2* (63)

822.6 X mass of nitric acid = 751.06 gram

b) % yields = ( 96.2%* 91.3% *91.4%)= 80.3%

3 0

3 years ago