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2 years ago

How many moles of O2 will be formed from 1.65 moles of KCIO3? (2KClO3–>2KCl + 3O2)

Chemistry

1 answer:

Oduvanchick [21]2 years ago

7 0

Answer:

2.475 mol of O2 formed.

Explanation:

Given 1.65 moles of KClO3 as the target amount in the reactant, used the coefficient of the balanced chemical reaction involved to determine the number of moles of O2 molecules formed.

x mole of O2 = 1.65 mol KClO3 x [(3 mol O2)/ (2 mol KClO3)] = 2.475 mol of O2

x mole of O2 formed = 2.475 mol of O2

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A metal sample is heated and placed into the water in a calorimeter at room temperature. Which statement best describes how the

andre [41]

Answer:

Energy transfers from the metal to the water and calorimeter until they are all at room temperature.

Explanation:

CHECK THE COMPLETE QUESTION BELOW;

A metal sample is heated and placed into the water in a calorimeter at room temperature. Which statement best describes how the calorimeter can be used to determine the specific heat capacity of the metal sample?

Energy transfers to the metal from the water and calorimeter until they are all at room temperature

. Energy transfers from the metal to the water and calorimeter until they are all at room temperature.

Energy transfers to the metal from the water and calorimeter until they all reach a single temperature.

Energy transfers from the metal to the water and calorimeter until they all reach a single temperature.

EXPLANATION;

Using calorimeter to determine the specific heat capacity of the metal sample can be associated to the theory of conservation of energy because heat which is a form of energy is been transfer of heat between the metal to the water and the calorimeter, this process will proceed till single temperature is attained.

The change in the amount of temperature of the water in the calorimeter is measured in order to get the difference in heat change of the calorimeter water.

CHANGE IN HEAT CAN BE CALCULATED USING THE FORMULA.

Q = cmΔT where Q is the change in heat , c is the specific heat capacity and ΔT is the change in temperature

6 0

2 years ago

If 36.9 mL of B2H6 reacted with excess oxygen gas, determine the actual yield of B2O3 if the percent yield of B2O3 was 75.7%. (T

zhuklara [117]

Answer: The actual yield of $B_2O_3$ is 60.0 g

Explanation:-

The balanced chemical reaction :

$B_2H_6(l)+3O_2(g)\rightarrow B_2O_3(s)+3H_2O(l)$

Mass of $B_2H_6$ = $Density\times Volume=1.131g/ml\times 36.9ml=41.7g$

$\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}$

$\text{Moles of} B_2H_6=\frac{41.7g}{27.668g/mol}=1.51moles$

According to stoichiometry:

1 mole of $B_2H_6$ gives = 1 mole of $B_2O_3$

1.51 moles of $B_2H_6$ gives = $\frac{1}{1}\times 1.51=1.51$ moles of $B_2O_3$

Theoretical yield of $B_2O_3=moles\times {Molar mass}}=1.14mol\times 69.62g/mol=79.3g$

Percent yield of $B_2O_3$ = $75.7\%$

$\%\text{ yield}=\frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100$

$75.7\%=\frac{\text{Actual yield}}{79.3}\times 100$

${\text{Actual yield}}=60.0g$

Thus the actual yield of $B_2O_3$ is 60.0 g

7 0

2 years ago

(b) When a sample of C2H5OH was combusted, the volume of CO2(g) produced was 18.0 L when measured at

myrzilka [38]

Answer:

i) The number of moles of CO₂ (g) produced from the reaction = 0.07663 mole

ii) The volume of C₂H₅OH (l), in mL, that was combusted to produce the volume of CO₂ (g)

collected = 2.234 mL

iii) The amount of heat, in KJ, that was released by the combustion reaction = 52.4 kJ

Explanation:

The balanced chemical reaction when ethanol is combusted is given as

C₂H₅OH (l) + 3O₂ (g) → 2CO₂ (g) + 3H₂O (g)

The volume of CO₂(g) produced was 18.0 L when measured at 21.7°C and 1.03 atm.

i) Number of moles of CO₂ (g) produced by the reaction

With the correct and logical assumption that CO₂ is an ideal gas, the ideal gas equation has the relation

PV = nRT

P = pressure = 1.03 atm = 1.03 × 101325 Pa = 10,435.96 Pa

V = Volume of the gas = 18.0 L = 0.018 m³

n = number of moles = ?

R = molar gas constant = 8.314 J/mol.K

T = absolute temperature in Kelvin = 21.7 + 273.15 = 294.85 K

(10,435.96 × 0.018) = n × 8.314 × 294.85

n = 0.076629106 = 0.07663 mole

ii) The volume of C₂H₅OH (l), in mL, that was combusted to produce the volume of CO₂(g)

collected.

Recall the stoichiometric balance of the reaction

C₂H₅OH (l) + 3O₂ (g) → 2CO₂ (g) + 3H₂O (g)

2 moles of CO₂ is obtained from 1 mole of C₂H₅OH

0.07663 mole of CO₂ will be obtained from (0.07663×1/2) mole of C₂H₅OH; that is, 0.03831 mole of C₂H₅OH.

But we can convert this number of moles used up to mass of C₂H₅OH produced

Mass = (Number of moles) × (Molar Mass)

Molar mass of C₂H₅OH = 46.07 g/mol

Mass of C₂H₅OH combusted from the reaction

= 0.03831 × 46.07 = 1.765 g

But density of C₂H₅OH = 0.79 g/mL

Density = (Mass)/(Volume)

Volume = (Mass)/(Density) = (1.765/0.79)

= 2.234 mL

iii) The amount of heat, in KJ, that was released by the combustion reaction.

The heat of combustion of C₂H₅OH at the temperature of the reaction = -1367.6 kJ/mol. (From literature)

1 mole of C₂H₅OH combusts to give 1367.6 kJ of heat

0.03831 mole of C₂H₅OH will give (0.03831×1367.6) = 52.39 kJ = 52.4 kJ

Hope this Helps!!!!

3 0

2 years ago

The combustion of 0.374 kg of methane in the presence of excess oxygen produces 0.983 kg of carbon dioxide. What is the percent

anyanavicka [17]

Assuming that the combustion formula is
CH4 + 2O2 --> 2H2O + CO2<span>,

That means for every 1 molecule of methane(CH4) there will be one molecule of carbon dioxide(</span>CO2) produced. Methane molecular weight 16, carbon dioxide molecular weight is 44. Then the percent yield should be:
1 * (0.374/ 16) /(0.983/44)= 0.374*44/ 0.983 * 16= 104.6%

You sure the number is correct? Percent yield should not exceed 100%

5 0

2 years ago

84. Heavy water, D2O (molar mass = 20.03 g mol–1), can be separated from ordinary water, H2O (molar mass = 18.01), as a result o

ipn [44]

Answer:

relative rate of diffusion is 1.05

Explanation:

According to Graham's law of difussion:

Rate of diffusion is inversely proportional to the square root of molecular weight of a molecule.

For two given molecules:

$\frac{(rate)_{1}}{(rate)_{2}}=\frac{\sqrt{M_{2}} }{\sqrt{M_{1}}}$

The given molecules are

Water = 18.01

Heavy water =20.03

Thus the relative rate of diffusion will be:

$\frac{(rate)_{water}}{(rate)_{heavywater}}=\sqrt{ \frac{20.03}{18.01}}=1.05$

8 0

2 years ago