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2 years ago

What volume of o2 is needed to react fully with 720. Ml of nh3

1 answer:

6 0

Ammonia undergoes combustion with oxygen to produce nitric oxide and water. The volume of the oxygen required to react with 720 ml of ammonia is 900 ml.

<h3>What is volume?</h3>

Volume is the area occupied by the substance and is the ratio of the mass to the density.

At STP, 1 mole of gas occupies 22.4 L of volume

Given,

Volume of ammonia reacted = 0.720 L

The combustion reaction is shown as,

$\rm 4NH_{3} + 5O_{2} \rightarrow 4NO +6H_{2}O$

From the stoichiometry of the reaction, it can be said that,

$(4 \times 22.4)$ L of ammonia reacts with $(5 \times 22.4)$ L of oxygen gas.

So, 0.720 L of ammonia will react with:

$\dfrac{ (5 \times 22.4)}{(4 \times 22.4)} \times 0.720 = 0.9 \;\rm L$

Therefore, the volume of oxygen required is 900 mL.

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A reaction produces 74.10 g Ca(OH)2 after 56.08 g CaO is added to 36.04 g H2O. How should the difference in the masses of reacta

melomori [17]

Cao + H2O ---->Ca(OH)2
Calculate the number of each reactant and the moles of the product
that is
moles = mass/molar mass
The moles of CaO= 56.08g/ 56.08g/mol(molar mass of Cao)= 1mole
the moles of water= 36.04 g/18 g/mol= 2.002moles
The moles of Ca (OH)2=74.10g/74.093g/mol= 1mole

The mass of differences of reactant and product can be therefore
explained as
1 mole of Cao reacted completely with 1 mole H2O to produce 1 mole of Ca(OH)2. The mass of water was in excess while that of CaO was limited

3 0

3 years ago

Be sure to answer all parts. Dimercaprol (HSCH2CHSHCH2OH) was developed during World War I as an antidote to arsenic-based poiso

Sauron [17]

Answer:

For A: The number of arsenic atoms are $3.4\times 10^{21}$

For B: The percent composition of mercury, thallium and chromium in their complexes are 61.76 %, 62.2 % and 29.51 % respectively.

Explanation:

For A:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of dimercaprol = 696 mg = 0.696 g (Conversion factor: 1 g = 1000 mg)

Molar mass of dimercaprol = 124.21 g/mol

Putting values in above equation, we get:

$\text{Moles of dimercaprol}=\frac{0.696g}{124.21g/mol}=0.0056mol$

According to mole concept:

1 mole of a compound contains $6.022\times 10^{23}$ number of molecules.

So, 0.0056 moles of dimercaprol will contain $0.0056\times 6.022\times 10^{23}=3.4\times 10^{21}$ number of molecules.

As, 1 molecule of dimercaprol binds with 1 atom of Arsenic

So, $3.4\times 10^{21}$ number of dimercaprol molecules will bind with = $1\times 3.4\times 10^{21}=3.4\times 10^{21}$ number of arsenic atoms

Hence, the number of arsenic atoms are $3.4\times 10^{21}$

For B:

We know that:

Molar mass of dimercaprol = 124.21 g/mol

Molar mass of mercury = 200.59 g/mol

Molar mass of thallium = 204.38 g/mol

Molar mass of chromium = 51.99 g/mol

Also, 1 molecule of dimercaprol binds with 1 metal atom.

To calculate the percentage composition of metal in a complex, we use the equation:

$\%\text{ composition of metal}=\frac{\text{Mass of metal}}{\text{Mass of complex}}\times 100$ ......(1)

For mercury:

Mass of Hg-complex = (200.59 + 124.21) = 324.8 g

Mass of mercury = 200.59 g

Putting values in equation 1, we get:

$\%\text{ composition of mercury}=\frac{200.59g}{324.8g}\times 100=61.76\%$

For thallium:

Mass of Tl-complex = (204.38 + 124.21) = 328.59 g

Mass of thallium = 204.38 g

Putting values in equation 1, we get:

$\%\text{ composition of thallium}=\frac{204.38g}{328.59g}\times 100=62.2\%$

For chromium:

Mass of Cr-complex = (51.99 + 124.21) = 176.2 g

Mass of chromium = 51.99 g

Putting values in equation 1, we get:

$\%\text{ composition of chromium}=\frac{51.99g}{176.2g}\times 100=29.51\%$

Hence, the percent composition of mercury, thallium and chromium in their complexes are 61.76 %, 62.2 % and 29.51 % respectively.

8 0

3 years ago

18.2L of gas at 95°C and 760 torr is placed in a 15L container at 80 degrees * C ; what is the new pressure ?

romanna [79]

Answer:

884.56 torr

Explanation:

Formula: $\frac{P_{1}V_{1} }{T_{1}} = \frac{P_{2}V_{2} }{T_{2}}$

P = Pressure

V = Volume

T = Temperature in kelvin (Celsius + 273.15)

$\frac{(760)(18.2) }{368.15} = \frac{P(15) }{353.15}$

$P = \frac{(760)(18.2)(353.15) }{(368.15)(15)}$

P = 884.56169

4 0

3 years ago

Calculate the value deltaG°

atroni [7]

Answer:

ΔG=ΔG0+RTlnQ where Q is the ratio of concentrations (or activities) of the products divided by the reactants. Under standard conditions Q=1 and ΔG=ΔG0 . Under equilibrium conditions, Q=K and ΔG=0 so ΔG0=−RTlnK . Then calculate the ΔH and ΔS for the reaction and the rest of the procedure is unchanged.

Explanation:

4 0

2 years ago

Select all that apply. Catalysts can save money by essentially lowering the:

k0ka [10]

Answer: Options (a) and (d) are the correct answer.

Explanation:

A catalyst is the substance which helps in increasing the rate of reaction.

Activation energy is the minimum amount of energy required by reactants to start the reaction. On addition of catalyst, the path of reaction changes because the energy barrier gap reduces and hence, the activation energy also decreases.

In the absence of catalyst, we need to increase the temperature so that reaction can occur quickly.

Whereas on addition of catalyst, there is no need to increase the temperature as the catalyst itself is sufficient to increase the rate of reaction. As a result, temperature should be lowered when there is addition of catalyst in the reaction.

Thus, we can conclude that catalysts can save money by essentially lowering the activation energy and temperature required.

4 0

3 years ago

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