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3 years ago

2KMnO4= K2MnO4+ MnO2+O2 how many grams of KMnO4 are required to produce 1.60 grams of O2

Chemistry

1 answer:

Sergeu [11.5K]3 years ago

5 0

Answer: 15.8 g of $KMnO_4$ will be required to produce 1.60 grams of $O_2$

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}$

$\text{Moles of} O_2=\frac{1.60g}{32g/mol}=0.05moles$

$2KMnO_4\rightarrow K_2MnO_4+MnO_2+O_2$

According to stoichiometry :

As 1 mole of $O_2$ is given by = 2 moles of $KMnO_4$

Thus 0.05 moles of $O_2$ is given by = $\frac{2}{1}\times 0.05=0.10moles$ of $KMnO_4$

Mass of $KMnO_4=moles\times {\text {Molar mass}}=0.10moles\times 158g/mol=15.8g$

Thus 15.8 g of $KMnO_4$ will be required to produce 1.60 grams of $O_2$

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Pure chlorobenzene is contained in a flask attached to an open-end mercury manometer. When the flask contents are at 58.3°C, the

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Answer:

The slope is 4661.4K

(B), the intercept is 18.156

The vapor pressure of chlorobenzene is 731.4mmHg

The percentage of chlorobenzene originally in the vapor that condenses is = 99.7%

Explanation:

Two sets of conditions (a and b) are observed for L1 and L2 at two different temperatures.

$T_{a}$ = 58.3°C

$L1_{a}$ = 747mmHg

$L2_{a}$ = 52mmHg

$T_{b}$ = 110°C

$L1_{a}$ = 577mmHg

$L2_{b}$ = 222mmHg

We need to convert the temperatures from Celsius to Kelvin (K)

$T_{a}$ = 58.3°C + 273.2

$T_{a}$ = 331.5k

$T_{b}$ = 110°C + 273.2

$T_{b}$ = 383.2k

We then calculate the vapor pressures of the chlorobenzene at each set of conditions by measuring the difference in the mercury levels.

$P^{0}$ = $P_{atm} - (P_1 -P_2)$

The vapor pressure under the first set of conditions is:

$P^{a}$ = 755mmHg - (747mmHg - 52mmHg)

$P^{a}$ = 60mmHg

The vapor pressure under the second set of conditions is:

$P^{b}$ = 755mmHg - (577mmHg = 222mmHg)

$P^{b}$ = 400mmHg

First question say we should find ΔH(slope) and B(intercept) in the Clausius- Clapeyron equation:

Using the Formula :

$In_p^{0} = \frac{- (delta) H}{RT} + B$

where (delta) H = ΔH

The slope of the $In_p^{0} = \frac{T_1T_2 In (P_2/P_1)}{T_1-T_2}$

Calculating the slope; we have:

$\frac{- (delta) H}{R} = \frac{T_1T_2 In (P_2/P_1)}{T_1-T_2}$

$\frac{- (delta) H}{R}$ = $\frac{331.5 * 383.2 * In (400/60)}{(331.5-383.2)}$

$\frac{- (delta) H}{R}$ = -4661.4k

$\frac{ (delta) H}{R}$ = 4661.4k

The intercept can be derived from the Clausius-Clayperon Equation by making B the subject of the formula. To calculate the intercept using the first set of condition above; we have:

B = $In_p_{1} + \frac{ (delta) H}{RT}$

B = $In 60 + \frac{4661.4}{331.5}$

B = 18.156

Thus the Clausius-Clayperon equation for chlorobenzene can be expressed as:

In P = $\frac{-4661.4k}{T} + 18.156$

In question (b), the air saturated with chlorobenzeneis 130°C, converting he temperature of 130°C to absolute units of kelvin(k) we have;

T = 130°C + 273.2

T = 403.2k

Calculating the vapor pressure using Clausius-Clapyeron equation: we have;

$In_p_{0}$ = $In_p_{0} = \frac{-4661.4}{403.2} + 18.156$

$In_p_{0}$ = 6.595

$p_{0}$ = $e^{6.595}$

$p_{0}$ = 731.mmHg

The vapor pressure of chlorobenzene is 731.4mmHg

The diagram of the flowchart with the seperate vapor and liquid stream can be found in the attached document below.

Afterwards, both the inlet and outlet conditions contain saturated liquid.

From the flowchart, the vapor pressure of chlorobenzene at the inlet and outlet temperatures are known:

$P_1(130^{0}C)$ = 731.44mmHg

$P_1(58.3^{0}C)$ = 60mmHg

To calculate the percentage of the chlorobenzene originally in the vapor pressure that condenses; we must first calculate the mole fractions of chlorobenzene for the vapor inlet and outlet using Raoult's Law:

$y_1 = \frac{P^0 (T)}{P}$

At inlet conditions, the mole fraction of chlorobenzene is:

$y_1 = \frac{731.44}{101.3}*\frac{101.3}{760}$

$y_1 = \frac{0.962 mol cholorobenzene}{mol saturated air}$

At outlet conditions , the mole fraction of chlorobenzene is:

$y_2 = \frac{60}{101.3}*\frac{101.3}{760}$

$y_2 = \frac{0.0789 mol cholorobenzene}{mol saturated air}$

Since there is no reaction, the total balance around the condensation is :

$n_1 = n_2 + n_3$

Let assume, that 100 moles of liquid chlorobenzene (CB) is condensed, therefore the equation becomes:

$n_1 = n_2$ + 100mol

The chlorobenzene balance using the mole fractions calculated above is :

$\frac{0.962mol CB}{mol(air)} * n_1 mol(air) = \frac{0.0789molCB}{mol(air)}*n_2mol(air) + 100CB$

substituting equation (1) into equation above; we have:

$\frac{0.962mol CB}{mol(air)} * n_1 mol(air) = \frac{0.0789molCB}{mol(air)}*(n_2-100)mol(air) + 100CB$

$0.962_n_1 mol = 0.0789_n_1mol + 100mol$

we can solve for n1, i.e ;

n1 = 104.3mol total air

Therefore the moles of chlorobenzene that will produce 100 moles of CB liquid is:

$n_C_B = 104.3 (moles) air * \frac{0.962 mol (CB)}{mol air}$

$n_C_B$ = 100.34mol CB

Now, calculating the percentage of chlorobenzene that condenses: we have;

% CB Condensation = $\frac{100mol}{100.34mol}*100%$

% CB Condensation = 99.7%

The percentage of chlorobenzene originally in the vapor that condenses is 99.7%

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