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3 years ago

What mass of H2 is needed to react with 8.75 g of O2 according to the following equation: O2(g) + H2(g) → H2O(g)?

Chemistry

2 answers:

alina1380 [7]3 years ago

4 0

Mass of H₂ needed to react with O₂ : 1.092 g

<h3>Further explanation</h3>

The concentration of a substance can be expressed in several quantities such as moles, percent (%) weight / volume,), molarity, molality, parts per million (ppm) or mole fraction. The concentration shows the amount of solute in a unit of the amount of solvent.

Reaction

O₂(g) + 2H₂(g) → 2H₂O(g)

mass of O₂ : 8.75 g

mol O₂(MW=32 g/mol) :

$\tt \dfrac{8.75}{32}=0.273$

From the equation, mol ratio of O₂ : H₂ = 1 : 2, so mol H₂ :

$\tt \dfrac{2}{1}\times 0.273=0.546$

Mass H₂ (MW=2 g/mol) :

$\tt 0.546\times 2=1.092~g$

Fittoniya [83]3 years ago

3 0

Answer:

Mass of H₂ needed to react with O₂ : 1.092 g

Explanation:

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3 years ago

Determine the equilibrium constant, Kp, for the following reaction, by using the two reference equations below: 2 NO(g) + O2(g)

klio [65]

Answer:

$Kp=3.07x10^6$

Explanation:

Hello,

In this case, by knowing the given reference reactions, one could rearrange them as follows:

$2 NO(g) \leftrightarrow N_2(g) + O_2(g); Kp_2 = \frac{1}{2.3 x 10^{-19}}=4.35x10^{18}$

$N_2(g) + 2O_2(g) \leftrightarrow 2NO_2(g);Kp_3=(8.4x10^{-7})^2=7.056x10^{-13}$

Subsequently, to obtain the main reaction, we add the aforementioned reference rearranged reactions as shown below (just as reference):

$2NO(g)+N_2(g)+2O_2\leftrightarrow 2NO_2(g)+N_2+O_2$

Consequently, the equilibrium constant is computed as:

$Kp=\frac{[N_2][O_2]}{[NO]^2} * \frac{[NO_2]^2}{[N_2][O_2]^2} =Kp_2*Kp_3=4.35x10^{18}*7.056x10^{-13}=3.07x10^6$

Best regards.

8 0

3 years ago

What are the various adverse effects of chemistry

Kisachek [45]

Answer:

Chemistry is the reaction between certain elements to create a new compound which may happen naturally but is mostly man made and hence has lot of good as well as adverse effects.

Explanation:

The product or compound that we get and the energy or outcome we receive is the ultimate result that decides whether chemical reaction that has happened is good or bad.

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4 0

3 years ago

A mixture of 0.10 mol of NO, 0.050 mol of H2, and 0.10 mol of H2O is placed in a 1.0- L vessel at 300 K . The following equilibr

umka2103 [35]

Answer:

The concentration of N2 at the equilibrium will be 0.019 M

Explanation:

Step 1: Data given

Number of moles of NO = 0.10 mol

Number of moles of H2 = 0.050 mol

Number of moles of H2O = 0.10 mol

Volume = 1.0 L

Temperature = 300K

At equilibrium [NO]=0.062M

Step 2: The balanced equation

2NO(g) + 2H2(g) → N2(g) + 2H2O(g)

Step 3: Calculate the initial concentration

Concentration = Moles / volume

[NO] = 0.10 mol / 1L = 0.10 M

[H2] = 0.050 mol / 1L = 0.050 M

[H2O] = 0.10 mol / 1L = 0.10 M

[N2] = 0 M

Step 4: Calculate the concentration at the equilibrium

[NO] at the equilibrium is 0.062 M

This means there reacted 0.038 mol (0.038M) of NO

For 2 moles NO we need 2 moles of H2 to produce 1 mol N2 and 2 moles of H2O

This means there will also react 0.038 mol of H2

The concentration at the equilibrium is 0.050 - 0.038 = 0.012 M

There will be porduced 0.038 moles of H2O, this means the final concentration pf H2O at the equilibrium is 0.100 + 0.038 = 0.138 M

There will be produced 0.038/2 = 0.019 moles of N2

The concentration of N2 at the equilibrium will be 0.019 M

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3 years ago

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Answer:

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