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3 years ago

5

Nitric acid is produced commercially by the Ostwald process, represented by the following equations. 4 NH3(g) + 5 O2(g) 4 NO(g)

+ 6 H2O(g) 2 NO(g) + O2(g) 2 NO2(g) 3 NO2(g) + H2O(l) 2 HNO3(aq) + NO(g) What mass in kg of NH3 must be used to produce 7.6 ✕ 106 kg HNO3 by the Ostwald process, assuming 100% yield in each reaction?

1 answer:

Marrrta [24]3 years ago

4 0

Answer:

The answer to your question is: 1538095.2 kg of NH3

Explanation:

MW HNO3 = 63 kg

MW NO2 = 46 kg

3 NO2(g) + H2O(l)--- 2 HNO3(aq) + NO(g)

3(46) kg-------------- 2(63) kg

x --------------- 7600000 kg

x = 7600000 x 138/126 = 8323809.5 kg og NO2

MW NO = 30

2 NO(g) + O2(g)---2 NO2(g)

2(30) ------------------2(46)

x ---------------- 8323809.5 kg

x = 8323809.5 x 60/92 = 5428571.4 kg of NO

MW NH3 = 17 kg

4 NH3(g) + 5 O2(g) 4 NO(g) + 6 H2O(g)

4(17) -------------------- 4(30)

x ----------------------- 5428571.4

x = 5428571.4 x 34 / 120

x = 1538095.2 kg of NH3

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over a 12.3 minuete period 5.13 E-3 moles of F2 gas effuses from a contaier. How many moles of CH4 gas could effuse from from th

Aleonysh [2.5K]

Answer : The moles of methane gas could be, $7.90\times 10^{-3}mol$

Solution :

According to the Graham's law, the rate of effusion of gas is inversely proportional to the square root of the molar mass of gas.

$R\propto \sqrt{\frac{1}{M}}$

or,

$(\frac{R_1}{R_2})=\sqrt{\frac{M_2}{M_1}}$

$[\frac{(\frac{n_1}{t_1})}{(\frac{n_2}{t_2})}]=\sqrt{\frac{M_2}{M_1}}$

where,

$R_1$ = rate of effusion of fluorine gas

$R_2$ = rate of effusion of methane gas

$n_1$ = moles of fluorine gas = $5.13\times 10^{-3}mol$

$n_2$ = moles of methane gas = ?

$t_1=t_2$ = time = 12.3 min (as per question)

$M_1$ = molar mass of fluorine gas = 38 g/mole

$M_2$ = molar mass of methane gas = 16 g/mole

Now put all the given values in the above formula 1, we get:

$[\frac{(\frac{5.13\times 10^{-3}mol}{12.3min})}{(\frac{n_2}{12.3min})}]=\sqrt{\frac{16g/mole}{38g/mole}}$

$n_2=7.90\times 10^{-3}mol$

Therefore, the moles of methane gas could be, $7.90\times 10^{-3}mol$

8 0

3 years ago

Magnesium metal (0.100 mol) and a volume of aqueous hydrochloric acid that contains 0.500 mol of HCl are combined and react to c

jok3333 [9.3K]

Answer:

2.24 L of hydrogen gas, measured at STP, are produced.

Explanation:

Given, Moles of magnesium metal, $Mg$ = 0.100 mol

Moles of hydrochloric acid, $HCl$ = 0.500 mol

According to the reaction shown below:-

$Mg_{(s)} + 2HCl_{(aq)}\rightarrow MgCl_2_{(aq)} + H_2_{(g)}$

1 mole of Mg reacts with 2 moles of $HCl$

0.100 mol of Mg reacts with 2*0.100 mol of $HCl$

Moles of $HCl$ must react = 0.200 mol

Available moles of $HCl$ = 0.500 moles

Limiting reagent is the one which is present in small amount. Thus, $Mg$ is limiting reagent.

The formation of the product is governed by the limiting reagent. So,

1 mole of $Mg$ on reaction forms 1 mole of $H_2$

0.100 mole of $Mg$ on reaction forms 0.100 mole of $H_2$

Mole of $H_2$ = 0.100 mol

At STP,

Pressure = 1 atm

Temperature = 273.15 K

Volume = ?

Using ideal gas equation as:

$PV=nRT$

where,

P is the pressure

V is the volume

n is the number of moles

T is the temperature

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

1 atm × V L = 0.100 × 0.0821 L.atm/K.mol × 273.15 K

<u>⇒V = 2.24 L</u>

2.24 L of hydrogen gas, measured at STP, are produced.

4 0

3 years ago

The ionic naming rules

natulia [17]

Identify and name the cation; this is a metal element or polyatomic cation.

Identify and name the anion; this is a nonmetal element. Change the suffix to '-ide,' or use the polyatomic anion name.

7 0

3 years ago

9. Given the following reaction:CO (g) + 2 H2(g) CH3OH (g)In an experiment, 0.45 mol of CO and 0.57 mol of H2 were placed in a 1

WITCHER [35]

Answer:

$Keq=11.5$

Explanation:

Hello,

In this case, for the given reaction at equilibrium:

$CO (g) + 2 H_2(g) \rightleftharpoons CH_3OH (g)$

We can write the law of mass action as:

$Keq=\frac{[CH_3OH]}{[CO][H_2]^2}$

That in terms of the change $x$ due to the reaction extent we can write:

$Keq=\frac{x}{([CO]_0-x)([H_2]_0-2x)^2}$

Nevertheless, for the carbon monoxide, we can directly compute $x$ as shown below:

$[CO]_0=\frac{0.45mol}{1.00L}=0.45M\\$

$[H_2]_0=\frac{0.57mol}{1.00L}=0.57M\\$

$[CO]_{eq}=\frac{0.28mol}{1.00L}=0.28M\\$

$x=[CO]_0-[CO]_{eq}=0.45M-0.28M=0.17M$

Finally, we can compute the equilibrium constant:

$Keq=\frac{0.17M}{(0.45M-0.17M)(0.57M-2*0.17M)^2}\\\\Keq=11.5$

Best regards.

3 0

4 years ago

Determine the frequency of radiation whose wavelength is 7.67 x 10^-7 cm. Then, determine the amount of energy in the radiation.

dimulka [17.4K]

Answer:

They frequency = 3.9 * 10^16 Hz

The amount of energy E = 2.58 *10^-17 J

Explanation:

Step 1: Data given

wavelength is 7.67 * 10^-7 cm

Step 2: Calculate the frequency

f = c / λ

⇒ with λ = the wavelength in nm = 7.67 nm

⇒ with c = the speed of light = 3.00 * 10^8 m/s

⇒ f = the frequency = TO BE DETERMINED

f = (3.00 * 10^8 m/s) / 7.67 * 10^-9 m

f = 3.9 * 10 ^16 /s = 3.9 * 10^16 Hz

They frequency is 3.9 * 10^16 Hz

Step 3: Calculate the amount of energy

E = h *f

⇒ with E = the amount of energy (in joule)

⇒ h = Planck's constant = 6.626 *10^-34 J*s

⇒ with f = the frequency

E = 6.626 *10^-34 J*s * 3.9 * 10^16 Hz

E = 2.58 *10^-17 J

8 0

3 years ago