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3 years ago

If 23.6 g of hydrogen gas reacts with 28.3 g of nitrogen gas, what is the maximum amount of product that can be produced?

Chemistry

1 answer:

Aleonysh [2.5K]3 years ago

8 0

Answer:

34.3 g NH3

Explanation:

M(H2) = 2*1 = 2 g/mol

M(N2) = 2*14 = 28 g/mol

M(NH3) = 14 + 3*1 = 17 g/mol

23.6 g H2* 1 mol/2 g = 11.8 mol H2

28.3 g N2 * 1 mol/28 g = 1.01 mol N2

3H2 + N2 ------> 2NH3

from reaction 3 mol 1 mol

given 11.8 mol 1.01 mol

We can see that H2 is given in excess, N2 is limiting reactant.

3H2 + N2 ------> 2NH3

from reaction 1 mol 2 mol

given 1.01 mol x

x = 2*1.01/1= 2.02 mol NH3

2.02 mol * 17g/1 mol ≈ 34.3 g NH3

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Which of these 2A elements has the largest atomic radius?

olya-2409 [2.1K]

Answer: Option (E) is the correct answer.

Explanation:

When we move from top to bottom in a group then there occurs an increase in atomic size of the atoms due to increase in the number of electrons.

For example, in group 2A elements beryllium is the smallest in size whereas radium being at the bottom is the largest in size.

Also, atomic number of beryllium is 4 and atomic number of radium is 88.

Thus, we can conclude that out of the given options radium is the 2A element which has the largest atomic radius.

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3 years ago

How do metals form chemical bonds with other metals?

gogolik [260]

Answer:

Metallic bonding

Explanation:

Metals have low ionization energies. Therefore, their valence electrons are easily delocalized (attracted to the neighbouring metal atoms). These delocalized electrons are then not associated with a specific metal atom. Since the electrons are “free”, the metal atoms have become cations, and the electrons are free to move throughout the whole crystalline structure.

We say that a metal consists of an array of cations immersed in a sea of electrons .

The electrons act as a “glue” holding the cations together.

Metallic bonds are the attractive forces between the metal cations and the sea of electrons.

In an NaK alloy, for example, the Na and K atoms contribute their valence electrons to the "sea". The atoms aren’t bonded to each other, but they are held in place by the metallic bonding.

4 0

2 years ago

Read 2 more answers

A 2.89 g sample of a metal chloride, MCl2, is dissolved in water and treated with excess aqueous silver nitrate. The silver chlo

dsp73

Answer:

58.0 g/mol

Explanation:

The reaction that takes place is:

MCl₂ + 2AgNO₃ → 2AgCl + M(NO₃)₂

First we calculate how many moles of silver chloride were produced, using its molar mass:

6.41 g AgCl ÷ 143.32 g/mol = 0.0447 mol AgCl

Then we convert AgCl moles into MCl₂ moles, using the stoichiometric ratio:

0.0447 mol AgCl * $\frac{1molMCl_2}{2molAgCl}$ = 0.0224 mol MCl₂

Now we calculate the molar mass of MCl₂, using the original mass of the sample:

2.86 g / 0.0224 mol = 127.68 g/mol

We can write the molar mass of MCl₂ as:

Molar Mass MCl₂ = Molar Mass of M + (Molar Mass of Cl)*2

127.68 g/mol = Molar Mass of M + (35.45 g/mol)*2

Finally we calculate the molar mass of M:

Molar Mass of M = 57 g/mol

The closest option is 58.0 g/mol.

6 0

2 years ago

Citric acid is often used as an acidity regulator in hot water canning of tomatoes. At 25°C it has Kat 7.4x10- and A.Hº = +4.1 k

gavmur [86]

Answer:

The rate constant at T = 100 C is 1.0*10⁻³

Explanation:

The Arrhenius equation relates two rate constants K1 and K2 measured at temperatures T1 and T2 as shown below:

$ln\frac{K_{2}}{K_{1}}=\frac{\Delta H^{0}rxn}{R}(\frac{1}{T_{1}}-\frac{1}{T_{2}})$

here, ΔHrxn = standard enthalpy change of the reaction

R = gas constant

From the given information:

K1 = 7.4*10^-4

T1 = 25 C = 25+273 = 298 K

T2 = 100 C = 100+273 = 373K

ΔH°=4.1kJ/mol

$ln\frac{K_{2}}{7.4*10^{-4}}=\frac{4.1 kJ/mol}{0.08314kJ/mol.K}(\frac{1}{298}-\frac{1}{373})K$

K2 = 1.03*10⁻³

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3 years ago

Is aluminum Pure or Impure?

mr_godi [17]

Answer:

The aluminium is impure

6 0

2 years ago