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3 years ago

what is the mass of water vapor produced when 3.2 liters reacts with 8.7 liters of oxygen gas at STP?

Chemistry

1 answer:

uranmaximum [27]3 years ago

8 0

Answer:

2.57g

Explanation:

First, let us write a balanced equation for the reaction. This is illustrated below:

2H2 + O2 —> 2H2O

Next let us determine the limiting reactant. This is achieved as follows:

From the equation,

2L H2 required 1L of O2.

Therefore, 3.2L of H will require = 3.2/2 = 1.6L of O2

From the calculation above, O2 is excess because the volume of O2 given from the question is far greater than the volume of O2 obtained from our calculation. Therefore, H2 is the limiting reactant.

Now let us covert 3.2L of H2 to mole. This is illustrated below:

1mole of a gas occupy 22.4L at stp

Therefore, Xmol of H2 will occupy 3.2L i.e

Xmol of H2 = 3.2/22.4 = 0.143mol

From the equation,

2moles of H2 produced 2moles of H2O.

Therefore, 0.143mol of H2 will also produce 0.143moles of H2O.

Now, we can obtain the mass of the water vapour produced by convert 0.143mol of H2O to gram. This is illustrated below:

Molar Mass of H2O = (2x1) + 16 = 2 + 16 = 18g/mol

Number of mole of H2O = 0.143mol

Mass of H2O =?

Mass = mole x Molar Mass

Mass of H2O = 0.143 x 18 = 2.57g

The mass of water vapour produce is 2.57g

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1 year ago

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3 years ago

If 16.00 g of O₂ reacts with 80.00 g NO, how many the excess reactant are left over? (enter only the value, round to whole numbe

pishuonlain [190]

Answer:

Explanation:

We will need a balanced equation with masses, moles, and molar masses of the compounds involved.

1. Gather all the information in one place with molar masses above the formulas and masses below them.

Mᵣ: 30.01 32.00 46.01

2NO + O₂ ⟶ 2NO₂

Mass/g: 80.00 16.00

2. Calculate the moles of each reactant

$\text{moles of NO} = \text{80.00 g NO} \times \dfrac{\text{1 mol NO}}{\text{30.01 g NO}} = \text{2.666 mol NO}\\\\\text{moles of O}_{2} = \text{16.00 g O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{32.00 g O}_{2}} = \text{0.5000 mol O}_{2}$

3. Calculate the moles of NO₂ we can obtain from each reactant

From NO:

The molar ratio is 2 mol NO₂:2 mol NO

$\text{Moles of NO}_{2} = \text{2.333 mol NO} \times \dfrac{\text{2 mol NO}_{2}}{\text{2 mol NO}} = \text{2.333 mol NO}_{2}$

From O₂:

The molar ratio is 2 mol NO₂:1 mol O₂

$\text{Moles of NO}_{2} = \text{0.5000 mol O}_{2}\times \dfrac{\text{2 mol NO}_{2}}{\text{1 mol Cl}_{2}} = \text{1.000 mol NO}_{2}$

4. Identify the limiting and excess reactants

The limiting reactant is O₂ because it gives the smaller amount of NO₂.

The excess reactant is NO.

5. Mass of excess reactant

(a) Moles of NO reacted

The molar ratio is 2 mol NO:1 mol O₂

$\text{Moles reacted} = \text{0.500 mol O}_{2} \times \dfrac{\text{2 mol NO}}{\text{1 mol O}_{2}} = \text{1.000 mol NO}$

(b) Mass of NO reacted

$\text{Mass reacted} = \text{1.000 mol NO} \times \dfrac{\text{30.01 g NO}}{\text{1 mol NO}} = \text{30.01 g NO}$

(c) Mass of NO remaining

Mass remaining = original mass – mass reacted = (80.00 - 30.01) g = 50 g NO

5 0

3 years ago

Compounds A, B, and C react according to the following equation. 3A(g) + 2B(g) 2C(g) At 100°C a mixture of these gases at equili

Vedmedyk [2.9K]

Answer:

The value of Kc for the reaction is 3.24

Explanation:

A reversible chemical reaction, indicated by a double arrow, occurs in both directions: reagents transforming into products ( direct reaction) and products transforming back into reagents (inverse reaction)

Chemical Equilibrium is the state in which direct and indirect reactions have the same reaction rate. Then taking into account the rate constant of a direct reaction and its inverse the chemical constant Kc is defined.

Being:

aA + bB ⇔ cC + dD

where a, b, c and d are the stoichiometric coefficients, the equilibrium constant with the following equation:

$Kc=\frac{[C]^{c} *[D]^{d} }{[A]^{a} *[B]^{b} }$

Kc is equal to the multiplication of the concentrations of the products raised to their stoichiometric coefficients divided by the multiplication of the concentrations of the reagents also raised to their stoichiometric coefficients.

Then, in the reaction 3A(g) + 2B(g) ⇔ 2C(g), the constant Kc is:

$Kc=\frac{[C]^{2} }{[A]^{3} *[B]^{2} }$

where:

[A]= 0.855 M
[B]= 1.23 M
[C]= 1.75 M

Replacing:

$Kc=\frac{1.75^{2} }{0.855^{3}*1.23^{2} }$

Solving you get:

Kc=3.24

<u><em>The value of Kc for the reaction is 3.24</em></u>

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