All categories

3 years ago

What volume of o2 at 874 mmhg and 33 ∘c is required to synthesize 18.0 mol of no?

Chemistry

2 answers:

Jlenok [28]3 years ago

8 0

<em>490.93 liters is required to synthesize 18.0 mol of NO </em>

<h3><em>Further explanation </em></h3>

Stokiometry in Chemistry learn about chemicals mainly emphasizes quantitative, such as the calculation of volume, mass, number, which is related to numbers, molecules, elements, etc.

A reaction coefficient is a number in the chemical formula of a substance involved in the reaction equation. The reaction coefficient is useful for equalizing reagents and products.

In the reaction there are also manifestations of reagent substances namely gas (g), liquid (liquid / l), solid (solid / s) and solution (aqueous / aq).

Reactions that occur:

4NH₃ (g) + 5O₂ (g) -> 4NO (g) + 6H₂O (g)

Comparison of coefficient = mole ratio

So that mole O₂ = 5/4 x mol NO

mole O₂ = 5/4 x 18

mole O₂ = 22.5

To find the volume, assuming the ideal gas, we use the formula:

$\large{\boxed{\boxed{\bold{PV=nRT}}}$

P = 874 mmhg = 874/760 mmhg = 1.15 atm

R = 0.082 L atm / mole K

T = 33 + 273 = 306 K

V = nRT / P

V = 22.5 0.082.306 / 1.15

V = 490.93 liters

<h3><em>Learn more </em></h3>

the ideal gas law

brainly.com/question/3637553

excess reactant

brainly.com/question/6857557

the percentage yield

brainly.com/question/12044319

limiting reactant

brainly.com/question/5798341

Keywords: ideal gas law, mole, NO, O₂

Nataly [62]3 years ago

3 0

1) Write the balanced chemical equation

4NH3 + 5O2 ----> 4NO + 6H2O

2) Write the molar ratios

4 mol NH3 : 5 mol O2 : 4 mol NO : 6 mol H2O

3) Write the proportions with the desired quantity of NO and the unknown quantity of O2

5 mol O2 / 4 mol NO = x / 18.0 mol NO

Solve for x: x = 18.0 mol NO * 5 mol O2 / 4 mol NO = 22.5 mol O2

4) Use the ideal gas equation to convert moles to volume

pV = nRT => V = nRT / p

n = 22.5 mol

R = 0.082 atm*liter / K * mol

T = 33 + 273.15 = 306.15 K

p = 874/760 atm = 1.15 atm

V = 22.5 mol * 0.082 atm*liter/K*mol * 306.15 K / 1.15 atm = 491.17 liter

Answer: 491 liters of O2

You might be interested in

The pressure exerted on a sample of gas is 735 mm Hg with a volume of 375 mL and a temperature of 25 degrees Celsius. What would

Darina [25.2K]

Answer:

= 498.13 mmHg

Explanation:

Using the combined gas law;

P1V1/T1 =P2V2/T2

in this case; P1 = 725 mmHg, V1 = 375 mL and T1 = 25°C + 273 = 298 K

P2 = ? V2 = 500 mL and T2 (standard temperature) = 0°C + 273 = 273 K

P2 = P1V1T2/T1V2

= (725 × 375 × 273)/(298 × 500)

<u>= 498.13 mmHg </u>

7 0

3 years ago

At 298 K, the rate constant for a reaction is 0.0346 s-1. What is the rate constant at 350K if the Ea = 50.2kJ/mol

frutty [35]

Answer:

0.702 /s

Explanation:

Rate constant at $[298 \mathrm{~K}, \mathrm{~K}_{1}=3.46 \times 10^{-2} \mathrm{~s}^{-1}$

Rate constant at $350 \mathrm{~K}, \mathrm{~K}_{2}=?$

$T_{1}=298 \mathrm{~K}$

$T_{2}=350 \mathrm{~K}$

Activation energy, $\mathrm{Ea}=50.2 \times 10^{3} \mathrm{~J} / \mathrm{mol}$

Use the following equation to calculate $K_{2}$ at $350 \mathrm{~K}$

$\ln \frac{\mathrm{K}_{2}}{\mathrm{~K}_{1}}=\frac{\mathrm{Ea}}{\mathrm{R}}\left[\frac{1}{\mathrm{~T}_{1}}-\frac{1}{\mathrm{~T}_{2}}\right]$

Therefore,

$\ln \left(\frac{K_{2}}{3.46 \times 10^{-2} \mathrm{~s}^{-1}}\right) &=\frac{50.2 \times 10^{3} \mathrm{~J} / \mathrm{mol}}{8.314 \mathrm{JK}^{-1} \mathrm{~mole}^{-1}}\left[\frac{1}{298 \mathrm{~K}}-\frac{1}{350 \mathrm{~K}}\right]$

$\ln \left(\frac{K_{2}}{3.46 \times 10^{-2} \mathrm{~s}^{-1}}\right) &=\frac{50.2 \times 10^{3} \mathrm{~J} / \mathrm{mol}}{8.314 \mathrm{JK}^{-1} \mathrm{~mole}^{-1}}\left[\frac{52 \mathrm{~K}}{298 \mathrm{~K} \times 350 \mathrm{~K}}\right]$