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3 years ago

Please help, its urgent.

Chemistry

1 answer:

san4es73 [151]3 years ago

5 0

Answer:

Mass of excess reactant left = 179.6 g

Limiting reactant = nitrogen

Mass of ammonia formed = 200.6 g

Explanation:

Given data:

Mass of nitrogen = 165.0 g

Mass of hydrogen = 215.0 g

Limiting reactant = ?

Mass of ammonia formed = ?

Mass of excess reactant left = ?

Solution:

Chemical equation:

N₂ + 3H₂ → 2NH₃

Number of moles of nitrogen:

Number of moles = mass/molar mass

Number of moles = 165.0 g/ 28 g/mol

Number of moles = 5.9 mol

Number of moles of hydrogen:

Number of moles = mass/molar mass

Number of moles = 215.0 g/ 2 g/mol

Number of moles = 107.5 mol

Now we will compare the moles of ammonia with both reactant.

H₂ : NH₃

3 : 2

107.5 : 2/3×107.5 = 71.7 mol

N₂ : NH₃

1 : 2

5.9 : 2/1×5.9 = 11.8 mol

Less number of moles of ammonia are formed by the nitrogen it will act as limiting reactant.

Mass of ammonia formed:

Mass = number of moles × molar mass

Mass = 11.8 mol × 17 g/mol

Mass = 200.6 g

Mass of hydrogen left:

We will compare the moles of hydrogen and nitrogen.

N₂ : H₂

1 : 3

5.9 : 3/1×5.9 = 17.7 mol

Out of 107.5 moles 17.7 moles of hydrogen react with nitrogen.

Number of moles left unreacted = 107.5 - 17.7 mol = 89.8 mol

Mass of hydrogen left:

Mass = number of moles × molar mass

Mass = 89.8 mol × 2 g/mol

Mass = 179.6 g

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satela [25.4K]

Density:

The density of a substance is equal to the ratio of its mass and volume. In general we can say the density of a substance is mass per unit volume.

Density = mass in g / volume in cubic cm or ml

Given that

Density = 0.940 g/ ml

Mass = 48.0 g

Therefore,

Volume = mass/ density

= 48.0 g/ 0.940 g/ ml

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Thus the correct answer is 51 ml

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3 years ago

Trial 12 1.Volume of Acid used ( mL) 2020 2.Molarity of acid used (M) 0.0160.016 3.Initial volume reading of base buret 018.6 4.

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Answer:

$V_B = 18.3mL$ -- Volume of base used

$M_B = 0.0175M$ --- Molarity of base

Explanation:

Given

$V_A = 20mL$ -- Volume of acid used

$V_B_1 = 18.6mL$ --- Buret Initial reading

$V_B_2 = 36.9mL$ --- Buret Final reading

$M_A = 0.016M$ --- Molarity of the acid

Solving (a): Volume of base used (VB)

This is calculated by subtracting the initial reading from the final reading of the base buret.

i.e.

$V_B = V_B_2 - V_B_1$

$V_B = 36.9mL - 18.6mL$

$V_B = 18.3mL$

Solving (b): Molarity base (MB)

This is calculated using:

$M_A * V_A = M_B * V_B$

Make MB the subject

$M_B = \frac{M_A * V_A }{V_B}$

This gives:

$M_B = \frac{0.016M *20mL}{18.3mL}$

$M_B = \frac{0.016M *20}{18.3}$

$M_B = \frac{0.32M}{18.3}$

$M_B = 0.0175M$

Solving (c): <em>There is no such thing as average molarity</em>

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3 years ago

A 2.50-l flask contains a mixture of methane (ch4) and propane (c3h8) at a pressure of 1.45 atm and 20°c. when this gas mixture

Cerrena [4.2K]

Answer:- Mole fraction of methane in the original gas mixture is 0.854.

Solution:- From given volume, pressure and temperature, we could calculate the total moles of the gaseous mixture of methane and propane using ideal gas law as:

PV = nRT

$n=\frac{PV}{RT}$

V = 2.50 L

P = 1.45 atm

T = 20 + 273 = 293 K

Let's plug in the values in the equation:

$n=(\frac{1.45*2.50}{0.0821*293})$

n = 0.151

Let's say the solution has X moles of methane. Then moles of propane would be = (0.151 - X)

The combustion equations of methane and propane are:

$CH_4+2O_2\rightarrow CO_2+2H_2O$

$C_3H_8+5O_2\rightarrow 3CO_2+4H_2O$

From methane balanced equation, there is 1:1 mol ratio between methane and carbon dioxide. So, X moles of methane would produce X moles of carbon dioxide.

From balanced equation of propane, there is 1:3 mol ratio between propane and carbon dioxide. So, (0.151 - X) moles of propane would give 3(0.151 - X) moles of carbon dioxide.

So, total moles of carbon dioxide that we would get from methane and propane combustion are:

X + 3(0.151 - X)

From given data, 8.60 g of carbon dioxide are formed by the combustion of gas mixture.

moles of Carbon dioxide = $8.60g(\frac{1mol}{44g})$