CHEMISTRY COLLEGE

In a lab experiment 80.0 g of ammonia [NH3] and 120 g of oxygen are placed in a reaction vessel. At the end of the reaction 72.2 g of water are obtained. Determine the percent yield of the reaction.

Answers

Answer 1

Answer:

The percent yield of the reaction : 89.14%

Further explanation

Reaction of Ammonia and Oxygen in a lab :

4 NH₃ (g) + 5 O₂ (g) ⇒ 4 NO(g)+ 6 H₂O(g)

mass NH₃ = 80 g

mol NH₃ (MW=17 g/mol):

$(80)/(17)=4.706$

mass O₂ = 120 g

mol O₂(MW=32 g/mol) :

$\tt (120)/(32)=3.75$

Mol ratio of reactants(to find limiting reatants) :

$\tt (4.706)/(4)/ (3.75)/(5)=1.1765/ 0.75\rightarrow O_2~limiting~reactant(smaller~ratio)$

mol of H₂O based on O₂ as limiting reactants :

mol H₂O :

$\tt (6)/(5)* 3.75=4.5$

mass H₂O :

4.5 x 18 g/mol = 81 g

The percent yield :

$\tt \%yield=(actual)/(theoretical)* 100\%\n\n\%yield=(72.2)/(81)* 100\%=89.14\%$

Related Questions

How many valence electrons must two atoms share to form a single covalent bond? answers A.2 B.4 C.3 D.1
What type of hybridization is needed to explain why ethyne, C2H2, is linear?
Which of the following statements is true about exothermic reactions?
Compare and contrast the characteristics of metals and nonmetals.
Calculate the moles of benzoic acid (C6H5COOH) in a 55.66 g sample of benzoic acid

Calculate the wavelength of A 75 kg athlete running a 7.0-minute mile

Answers

Answer:

$\lambda =2.31x10^(-36)m$

Explanation:

Hello,

In this case, since the Broglie's wavelength for bodies is defined via:

$\lambda =(h)/(mv)$

Whereas h accounts for the Planck's constant, m the mass and v the velocity, which is:

$v=(1mile)/(7.0min)*(1609.34m)/(1mile)*(1min)/(60s)=3.83(m)/(s)$

Thus, the wavelength turns out:

$\lambda =(6.63x10^(-34)kg(m^2)/(s) )/(75kg*3.83(m)/(s) ) \n\n\lambda =2.31x10^(-36)m$

Best regards.

PLEASE HELP ASAP‼️ Which of the following elements has the highest metallic characterA. Zt
B. Rh
C. Hf
D. Co

Answers

Answer: C

Explanation:

Excess protons in the blood decrease the amount of HCO − 3 and thus reduce the buffering capacity of blood. A rapid drop in pH could lead to death. Normal values for blood are pH = 7.4 , [ HCO − 3 ] = 24.0 mM , [ CO 2 ] = 1.2 mM . (a) If a patient has a blood pH = 7.03 and [CO2] = 1.2 mM, what is the [HCO3−] in the patient’s blood? The pKa of HCO3−= 6.1.(b) Suggest a possible treatment for metabolic acidosis.

(c) Why might the suggestion for part (b) be of benefit to middle-distance runners?

Answers

Answer:

a) [HCO₃⁻] = 10,2 mM.

b) Sodium bicarbonate.

c) Yes.

Explanation:

a) The equilibrium of this reaction is:

CO₂ + H₂O ⇄ HCO₃⁻ + H⁺

Using Henderson-Hasselbalch equation:

pH = pka + log₁₀ $([HCO_(3)^-])/([CO_2])$

Replacing:

7,03 = 6,1 + log₁₀ $([HCO_(3)^-])/([1,2 mM])$

Thus, [HCO₃⁻] = 10,2 mM

b) A possible treatment of metabolic acidosis is with sodium bicarbonate. By Le Chateleir's principle the increasing of HCO₃⁻ will shift the equilibrium to the left decreasing thus, H⁺ concentration.

c) The shifting of the equilibrium to the left will increase CO₂ concentration producing in the body the need to increase breathing, increasing, thus, concentration of O₂ improving cardiac function in exercise.

What is the molarity of a solution prepared from 25.0 grams of methanol (CH3OH, density = 0.792 g/mL) with 100.0 milliliters of ethanol (CH3CH2OH)? Assume the volumes are additive.

Answers

Final answer:

The molarity of a solution prepared from 25.0 grams of methanol and 100.0 milliliters of ethanol is approximately 7.80 M.

Explanation:

This is a question about calculating molarity, which is a measure of concentration using moles per liter. To calculate the molarity of a methanol in ethanol, we first have to convert the mass of methanol into moles. The molar mass of methanol (CH3OH) is about 32.04 g/mol. Therefore, 25.0 g of methanol equals about 0.780 moles (25.0 g ÷ 32.04 g/mol).

Next, the volume of ethanol needs to be converted from milliliters to liters. Thus, 100.0 mL becomes 0.100 L. Finally, the molarity is calculated by dividing the moles of methanol by the volume of the ethanol in liters, resulting in a molarity of approximately 7.80 M (0.780 moles ÷ 0.100 L).

Learn more about Molarity here:

brainly.com/question/8732513

#SPJ12

Suppose you have been given the task of distilling a mixture of hexane + toluene. Pure hexane has a refractive index of 1.375 and pure toluene has a refractive index of 1.497. You collect a distillate sample which has a refractive index of 1.441. Assuming that the refractive index of the hexane + toluene mixture varies linearly with mole fraction, what is the mole fraction of hexane in your sample?

Answers

Answer:

0.4590

Explanation:

How the refractive index of the hexane + toluene mixture varies linearly with mole fraction, it means that the mole fraction is the fraction that each pure index contribute for the mixture index, so, calling xh the mole fraction of hexane and xt the mole fraction of toluene:

1.375xh + 1.497xt = 1.441

And, xh + xt = 1 (because there are only hexane and toluene in the mixture), so xt = 1- xh

1.375xh + 1.497(1-xh) = 1.441

1.375xh + 1.497 - 1.497xh = 1.441

-0.122xh = -0.056

xh = -0.056/(-0.122)

xh = 0.4590

Combustion of Solid Fuel. A fuel analyzes 74.0 wt % C and 12.0% ash (inert). Air is added to burn the fuel, producing a flue gas of 12.4% CO2, 1.2% CO, 5.7% O2, and 80.7% N2. Calculate the kg of fuel used for 100 kg mol of outlet flue gas and the kg mol of air used. (Hint: First calculate the mol O2 added in the air, using the fact that the N2 in the flue gas equals the N2 added in the air. Then make a carbon balance to obtain the total moles of C added.)

Answers

Answer:

Kg of fuel used = 220.54Kg

Explanation:

The concept of material balance is applied here by taking into consideration the percentage composition of each flue gas in the atmosphere. This is used with each of the percentage composition of the flue gas that burn in the fuel and the detailed steps is as shown in the attachment.

Other Questions

the reaction A->B is second order in A and the rate constant is 0.039L/mol s. if it took 23 sec for the concentration of A to decrease to 0.30 M, what was the starting concentration of A

If a cell is 80% water and the outside environment is 90% water. What is likely to happen? A. water will rush into the cell B. net movement of water will be equal C. water will not move into or out of the cell D. water will rush out of the cell

WORTH A LOT OF POINTS! just copy what on the picture for notes so i can copy and paste i do not feel like writing all of that down

1. An isotope of cesium-137 has a half-life of 30 years. If 5.0 g of cesium-137 decays over 60 years, how many grams will remain?