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2 years ago

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Calculate the number of millimoles contained in 500mg of FeSO4•C2H4(NH3)2SO4.4H20 Calculate the number of grams of BaCrO4 that w

ould have to be dissolved and diluted to 100ml to prepare a 0.200M solution.

1 answer:

VikaD [51]2 years ago

4 0

Answer:

$Millimoles\ of\ FeSO_4.C_2H_4(NH_3)_2SO_4.4H_2O= 1.3084\ millimoles$

The mass of $BaCrO_4$ required = 5.0674 g

Explanation:

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Given :

Mass = 500 mg

Molar mass of $FeSO_4.C_2H_4(NH_3)_2SO_4.4H_2O$ = 382.1455 g/mol

Thus,

$Millimoles= \frac{500\ mg}{382.1455\ g/mol}$

$Millimoles= 1.3084\ millimoles$

Also,

Molar mass of $BaCrO_4$ = 253.37 g/mol

Let the mass of the salt to prepare 100 mL 0.200 M solution = x g

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{x\ g}{253.37\ g/mol}$

$Moles= \frac{x}{253.37}\ mol$

Given that volume = 100 mL

Also,

$1\ mL=10^{-3}\ L$

So, Volume = 100 / 1000 L = 0.1 L

Molarity = 0.200 M

Considering:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

$0.200=\frac{x}{253.37\times {0.1}}$

x = 5.0674 g

<u>The mass of $BaCrO_4$ required = 5.0674 g</u>

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Answer:

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Explanation:

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That means, in therms of moles of oxygen we can write:

0.9050 moles = 2X/16.04 + 13/2Y/ 58.12

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<em>Where 16.04 and 58.12 are molar masses of methane and butane</em>

That is because if X is the mass of methane:

X g Methane * (1mol / 16.04g) = Moles methane

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Replacing (1) in (2):

0.9050 = 0.12469X + 0.11184 (7.519 - X)

0.9050 = 0.12469X + 0.841 - 0.11184X

0.0641 = 0.01285X

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The reaction SO2(g)+2H2S(g)←→3S(s)+2H2O(g) is the basis of a suggested method for removal of SO2 from power-plant stack gases. T

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Explanation :

The given balanced chemical reaction is,

$SO_2(g)+2H_2S(g)\rightleftharpoons 3S(s)+2H_2O(g)$

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

$\Delta G^o=G_f_{product}-G_f_{reactant}$

$\Delta G^o=[n_{S(s)}\times \Delta G_f^0_{(S(s))}+n_{H_2O(g)}\times \Delta G_f^0_{(H_2O(g))}]-[n_{SO_2(g)}\times \Delta G_f^0_{(SO_2(g))}+n_{H_2S(g)}\times \Delta G_f^0_{(H_2S(g))}]$

where,

$\Delta G^o$ = standard free energy of reaction = ?

n = number of moles

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

$\Delta G^o=[3mole\times (0kJ/mol)+2mole\times (-228.57kJ/mol)]-[1mole\times (-300.4kJ/mol)+2mole\times (-33.01kJ/mol)]$

$\Delta G^o=-90.72kJ/mol$

Now we have to calculate the value of $K_p$

$\Delta G^o=-RT\ln K_p$

where,

$\Delta G_^o$ = standard Gibbs free energy = -90.72 kJ/mol

R = gas constant = 8.314 J/mole.K

T = temperature = 298 K

$K_p$ = equilibrium constant = ?

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

$-90.72kJ/mol=-(8.314J/mol.K)\times (298K) \ln K_p$

$K_p=7.98\times 10^{15}$

Now we have to calculate the value of $K_p$ .

The given balanced chemical reaction is,

$SO_2(g)+2H_2S(g)\rightleftharpoons 3S(s)+2H_2O(g)$

The expression for equilibrium constant will be :

$K_p=\frac{(p_{H_2O})^2}{(p_{H_2S})^2\times (p_{SO_2})}$

In this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted.

Let the equilibrium $SO_2$ pressure be, x

Pressure of $SO_2$ = Pressure of $H_2S$ = x

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

$7.98\times 10^{15}=\frac{(22)^2}{(x)^2\times (x)}$

$x=3.93\times 10^{-5}torr$

Thus, the equilibrium $SO_2$ pressure is, $3.93\times 10^{-5}torr$

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