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3 years ago

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A chemist adds 220.0ml of a 0.194 m potassium permanganate kmno4 solution to a reaction flask. calculate the mass in grams of po

tassium permanganate the chemist has added to the flask

1 answer:

Tema [17]3 years ago

5 0

The formula of determining the molarity is:

molarity = $\frac{moles of solute}{volume of solution in liters}$ -(1)

Volume of solution = $220.0 mL$ (given)

Since, 1 mL = 0.001 L

Therefore, volume of the solution = 0.22 L

Molarity = 0.194 M (given)

Substituting the values in formula (1)

$0.194 = \frac{moles of solute}{0.22}$

$moles of solute= 0.22 L\times 0.194 mole/L = 0.0427 mole$

Now, number of moles = $\frac{mass}{Molar mass}$ -(2)

Molar mass of $KMnO_4$ = $39.0983+54.938+4\times 15.99 = 158.0323 g/mol$

Substituting the value of number of moles and molar mass of $KMnO_4$ in equation -(2)

$0.0427 mole = \frac{mass}{158.0323 g/mole}$

$mass = 158.0323 g/mole\times 0.0427 mole = 6.748 g$

Hence, the mass in grams of potassium permanganate the chemist has added to the flask is $6.748 g$ .

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Answer: 4.5 moles of $H_{2}$ can be made from complete reaction of 3.0 moles of Al.

Explanation:

The given reaction equation is as follows.

$2Al + 6HCl \rightarrow 2AlCl_{3} + 3H_{2}$

This shows that 2 moles of Al reacts with 6 moles of HCl. So, the amount of HCl required to react with 1 mole Al is three times the amount of HCl.

Therefore, 3 moles of Al will react with 9 moles of HCl to give 3 moles of $AlCl_{3}$ and $\frac{9}{2}$ moles of $H_{2}$ .

The reaction equation now will be as follows.

$3Al + 9HCl \rightarrow 3AlCl_{3} + \frac{9}{2}H_{2}$

The moles $\frac{9}{2}$ can also be written as 4.5 moles.

Thus, we can conclude that 4.5 moles of $H_{2}$ can be made from complete reaction of 3.0 moles of Al.

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3 years ago

How many moles of sand (SiO2) are in 30 g of sand?

anastassius [24]

<h3>Answer:</h3>

0.50 mol SiO₂

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<u>Step 1: Define</u>

30 g SiO₂ (sand)

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<u>Step 3: Convert</u>

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3 years ago

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Use the given data at 500 K to calculate ΔG°for the reaction

Anton [14]

Answer : The value of $\Delta G^o$ for the reaction is -959.1 kJ

Explanation :

The given balanced chemical reaction is,

$2H_2S(g)+3O_2(g)\rightarrow 2H_2O(g)+2SO_2(g)$

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

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

$\Delta H^o=[n_{H_2O}\times \Delta H_f^0_{(H_2O)}+n_{SO_2}\times \Delta H_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta H_f^0_{(H_2S)}+n_{O_2}\times \Delta H_f^0_{(O_2)}]$

where,

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

n = number of moles

$\Delta H_f^0$ = standard enthalpy of formation

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

$\Delta H^o=[2mole\times (-242kJ/mol)+2mole\times (-296.8kJ/mol)}]-[2mole\times (-21kJ/mol)+3mole\times (0kJ/mol)]$

$\Delta H^o=-1035.6kJ=-1035600J$

conversion used : (1 kJ = 1000 J)

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

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

$\Delta S^o=[n_{H_2O}\times \Delta S_f^0_{(H_2O)}+n_{SO_2}\times \Delta S_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta S_f^0_{(H_2S)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]$

where,

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

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

$\Delta S^o=[2mole\times (189J/K.mol)+2mole\times (248J/K.mol)}]-[2mole\times (206J/K.mol)+3mole\times (205J/K.mol)]$

$\Delta S^o=-153J/K$

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 500 K.

$\Delta G^o=(-1035600J)-(500K\times -153J/K)$

$\Delta G^o=-959100J=-959.1kJ$

Therefore, the value of $\Delta G^o$ for the reaction is -959.1 kJ

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Oksanka [162]

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Ulleksa [173]

Answer:

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