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Effectus [21]
2 years ago
8

Question 17 In the Haber reaction, patented by German chemist Fritz Haber in 1908, dinitrogen gas combines with dihydrogen gas t

o produce gaseous ammonia. This reaction is now the first step taken to make most of the world's fertilizer. Suppose a chemical engineer studying a new catalyst for the Haber reaction finds that 786. liters per second of dinitrogen are consumed when the reaction is run at 222.°C and 0.35atm. Calculate the rate at which ammonia is being produced. Give your answer in kilograms per second. Round your answer to 2 significant digits.
Chemistry
1 answer:
mars1129 [50]2 years ago
5 0

Answer:

Explanation:

N₂       + 3H₂     =     2 NH₃

1 vol                         2 vol

786 liters               1572 liters

786 liters of dinitrogen will result in the production of 1572 liters of ammonia

volume of ammonia V₁ = 1572 liters

temperature T₁ = 222 + 273 = 495 K

pressure = .35 atm

We shall find this volume at NTP

volume V₂ = ?

pressure = 1 atm

temperature T₂ = 273

\frac{P_1V_1}{T_1} =\frac{P_2V_2}{T_2}

\frac{.35\times 1572}{495} =\frac{1\times V_2}{ 273 }

V_2 =303.44 liter .

mol weight of ammonia = 17

At NTP mass of 22.4 liter of ammonia will have mass of 17 gm

mass of 303.44 liter of ammonia will be equal to (303.44 x 17) / 22.4 gm

= 230.28 gm

=.23 kg / sec .

Rate of production of ammonia = .23 kg /s .

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Answer:

The molecular formula of the compound is C_{7}H_{6}O_{2}.

Explanation:

Let consider that given percentages are mass percentages, so that mass of each element are determined by multiplying molar massof the organic acid by respective proportion. That is:

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m_{C} = \frac{68.84}{100}\times \left(122.12\,\frac{g}{mol} \right)

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Hydrogen

m_{H} = \frac{4.96}{100}\times \left(122.12\,\frac{g}{mol} \right)

m_{H} = 6.057\,g

Oxygen

m_{O} = \frac{26.20}{100}\times \left(122.12\,\frac{g}{mol} \right)

m_{O} = 31.995\,g

Now, the number of moles (n), measured in moles, of each element are calculated by the following expression:

n = \frac{m}{M}

Where:

m - Mass of the element, measured in grams.

M- Molar mass of the element, measured in grams per mol.

Carbon (m_{C} = 84.067\,g, M_{C} = 12.011\,\frac{g}{mol})

n = \frac{84.067\,g}{12.011\,\frac{g}{mol} }

n = 7

Hydrogen (m_{H} = 6.057\,g, M_{H} = 1.008\,\frac{g}{mol})

n = \frac{6.057\,g}{1.008\,\frac{g}{mol} }

n = 6

Oxygen (m_{O} = 31.995\,g, M_{O} = 15.999\,\frac{g}{mol})

n = \frac{31.995\,g}{15.999\,\frac{g}{mol} }

n = 2

For each mole of organic acid, there are 7 moles of carbon, 6 moles of hydrogen and 2 moles of oxygen. Hence, the molecular formula of the compound is:

C_{7}H_{6}O_{2}

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Answer:

Excess=3.53g

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

2Mg(s)+O_2(g) \rightarrow 2MgO(s)

Next, we identify the limiting reactant by computing the moles of magnesium oxide yielded by 3.86 g of magnesium and 155 mL of oxygen at the given conditions via their 2:1:2 mole ratios and the ideal gas equation:

n_{MnO}^{by \ Mg}=3.86gMg*\frac{1molMg}{24.3gMg}*\frac{2molMgO}{2molMg}  =0.159molMgO\\\\n_{MnO}^{by \ O_2}=\frac{1atm*0.155L}{0.082\frac{atm*L}{molO_2*K}*275K} *\frac{2mol MgO}{1molO_2} =0.0137molMgO

It means that the limiting reactant is the oxygen as it yields the smallest amount of magnesium oxide. Next, we compute the mass of magnesium consumed the oxygen only:

m_{Mg}^{consumed}=0.0137molMgO*\frac{2molMg}{2molMgO} *\frac{24.3gMg}{1molMg} =0.334gMg

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Answer:

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Explanation:

The law of definite proportion states that the chemical composition by mass of a chemical compound is always constant. For instance, a chemical compound that is made up of two elements will always contain the same proportions of the constituent elements regardless of the quantity of chemical that was used.

Using the law of definite proportion, we can determine the proportion of sodium and chlorine in NaCl and the proportion of potassium and chlorine in KCl if the mass of chlorine that was used is known. Based on the results obtained, one can easily determine the mixtures that has higher proportion of KCl.

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(b) Concentration of PO₄³⁻

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