Solution 1
2 months ago
4.17 moles. Good luck! :)
📚 Related Questions
Question
When copper reacts with silver nitrate according to the equation, the number of grams of copper required to produce 432 grams of silver is -?
Solution 1
Answer is: mass of copper is 127 grams.
Balanced chemical reaction: Cu(s) + 2AgNO₃(aq) → Cu(NO₃)₂(aq) + 2Ag(s).
m(Ag) = 432 g.
n(Ag) = m(Ag) ÷ M(Ag).
n(Ag) = 432 g ÷ 108 g/mol.
n(Ag) = 4 mol.
From chemical reaction: n(Ag) : n(Cu) = 2 : 1.
n(Cu) = 4 mol ÷ 2 = 2 mol.
m(Cu) = n(Cu) · M(Cu).
m(Cu) = 2 mol · 63.5 g/mol.
m(Cu) = 127 g.
Balanced chemical reaction: Cu(s) + 2AgNO₃(aq) → Cu(NO₃)₂(aq) + 2Ag(s).
m(Ag) = 432 g.
n(Ag) = m(Ag) ÷ M(Ag).
n(Ag) = 432 g ÷ 108 g/mol.
n(Ag) = 4 mol.
From chemical reaction: n(Ag) : n(Cu) = 2 : 1.
n(Cu) = 4 mol ÷ 2 = 2 mol.
m(Cu) = n(Cu) · M(Cu).
m(Cu) = 2 mol · 63.5 g/mol.
m(Cu) = 127 g.
Question
How does the total volume of gas particles compare to the volume of the space between gas particles?
Solution 1
Answer:
There is more space between gas particles than the size of the particles.
Explanation:
This scenario can be understand by taking a very simple example. As we know that 1 mole of any gas at standard temperature and pressure occupy 22.4 liters of volume. Lets take Hydrogen gas and Oxygen gas, 1 mole of each gas will occupy same volume. Why it is so? Why same volume although Oxygen is 16 times more heavier? This is because the space between gas molecules is very large. Approximately the distance between gas molecules is 300 times greater than their own diameter from its neighbor molecules.
There is more space between gas particles than the size of the particles.
Explanation:
This scenario can be understand by taking a very simple example. As we know that 1 mole of any gas at standard temperature and pressure occupy 22.4 liters of volume. Lets take Hydrogen gas and Oxygen gas, 1 mole of each gas will occupy same volume. Why it is so? Why same volume although Oxygen is 16 times more heavier? This is because the space between gas molecules is very large. Approximately the distance between gas molecules is 300 times greater than their own diameter from its neighbor molecules.
Question
Gas stoichiometry: what volume of oxygen at 25 degrees celsius and 1.04 atm is needed for the complete combustion of 5.53 grams of propane?
Solution 1
Answer is: volume of oxygen is 14.7 liters.
Balanced chemical reaction: C₃H₈ + 5O₂ → 3CO₂ + 4H₂O.
m(C₃H₈-propane) = 5.53 g.
n(C₃H₈) = m(C₃H₈) ÷ M(C₃H₈).
n(C₃H₈) = 5.53 g ÷ 44.1 g/mol.
n(C₃H₈) = 0.125 mol.
From chemical reaction: n(C₃H₈) : n(O₂) = 1 : 5.
n(O₂) = 0.625 mol.
T = 25° = 298.15K.
p = 1.04 atm.
R = 0.08206 L·atm/mol·K.
Ideal gas law: p·V = n·R·T.
V(O₂) = n·R·T / p.
V(O₂) = 0.625 mol · 0.08206 L·atm/mol·K · 298.15 K / 1.04 atm.
V(O₂) = 14.7 L.
Balanced chemical reaction: C₃H₈ + 5O₂ → 3CO₂ + 4H₂O.
m(C₃H₈-propane) = 5.53 g.
n(C₃H₈) = m(C₃H₈) ÷ M(C₃H₈).
n(C₃H₈) = 5.53 g ÷ 44.1 g/mol.
n(C₃H₈) = 0.125 mol.
From chemical reaction: n(C₃H₈) : n(O₂) = 1 : 5.
n(O₂) = 0.625 mol.
T = 25° = 298.15K.
p = 1.04 atm.
R = 0.08206 L·atm/mol·K.
Ideal gas law: p·V = n·R·T.
V(O₂) = n·R·T / p.
V(O₂) = 0.625 mol · 0.08206 L·atm/mol·K · 298.15 K / 1.04 atm.
V(O₂) = 14.7 L.
Question
A gas occupies 2240.0 l at 373 k. what are the volumes at standard temperature answers
Solution 1
Answer is: the volumes at standard temperature is 1639.46 L.
V₁(gas) = 2240.0 L.
T₁(gas) = 373 K.
T₂(gas) = 273 K, standard temperature.
V₂(gas) = ?
Charles' Law: The Temperature-Volume Law - the volume of a given amount of gas held at constant pressure is directly proportional to the Kelvin temperature:
V₁/T₁ = V₂/T₂.
2240 L/373 K = V₂/273 K.
V₂ = 1639.46 L.
V₁(gas) = 2240.0 L.
T₁(gas) = 373 K.
T₂(gas) = 273 K, standard temperature.
V₂(gas) = ?
Charles' Law: The Temperature-Volume Law - the volume of a given amount of gas held at constant pressure is directly proportional to the Kelvin temperature:
V₁/T₁ = V₂/T₂.
2240 L/373 K = V₂/273 K.
V₂ = 1639.46 L.
Question
Question 13 unsaved the nonmetals in groups 5a, 6a, and 7a: question 13 options: lose electrons when they form ions. form ions with charges of 3-, 2-, and 1-, respectively. form positively charged ions. form ions with a numerical charge equal to their group number.
Solution 1
How does the law of conservation of mass apply to this reaction: C2H4 + O2 → H2O + CO2?
Question
Question 3 what geometric arrangement of charge clouds is expected for an atom that has five charge clouds? trigonal bipyramidal square planar octahedral tetrahedral
Solution 1
AX5E0 - trigonal bipyramidal - zero lone pairs;
AX4E1 - seesaw - 1 lone pair;
AX3E2 - T-shaped - 2 lone pairs;
AX2E3 - linear - 3 lone pairs;
The only option that matches is trigonal bipyramidal.
Question
Question 10 unsaved in a chemical reaction, the mass of the products question 10 options: has no relationship to the mass of the reactants. is less than the mass of the reactants. is greater than the mass of the reactants. is equal to the mass of the reactants.
Solution 1
What are youre answer choices????
Question
What volume of O2( g. at 810. mmHg pressure is required to react completely with a 4.50g sample of C(s) at 48°C? 2 C(s) + O2(
g. → 2 CO(
g.
Solution 1
Answer is: volume of oxygen is 4.63 liters.
Balanced chemical reaction: 2C + O₂ → 2CO.
m(C) = 4.50 g.
n(C) = m(C) ÷ M(C).
n(C) = 4.50 g ÷ 12 g/mol.
n(C) = 0.375 mol.
From chemical reaction: n(C) : n(O₂) = 2 : 1.
n(O₂) = 0.1875 mol.
T = 48°C = 321.15 K.
p = 810 mmHg ÷ 760 mmHg/atm= 1.066 atm.
R = 0.08206 L·atm/mol·K.
Ideal gas law: p·V = n·R·T.
V(O₂) = n·R·T / p.
V(O₂) = 0.1875 mol · 0.08206 L·atm/mol·K · 321.15 K / 1.066 atm.
V(O₂) = 4.63 L.
Balanced chemical reaction: 2C + O₂ → 2CO.
m(C) = 4.50 g.
n(C) = m(C) ÷ M(C).
n(C) = 4.50 g ÷ 12 g/mol.
n(C) = 0.375 mol.
From chemical reaction: n(C) : n(O₂) = 2 : 1.
n(O₂) = 0.1875 mol.
T = 48°C = 321.15 K.
p = 810 mmHg ÷ 760 mmHg/atm= 1.066 atm.
R = 0.08206 L·atm/mol·K.
Ideal gas law: p·V = n·R·T.
V(O₂) = n·R·T / p.
V(O₂) = 0.1875 mol · 0.08206 L·atm/mol·K · 321.15 K / 1.066 atm.
V(O₂) = 4.63 L.
Question
What is the purpose of the catalyst? A. to start the chemical reaction
B. to stop the chemical reaction
C. to speed up the chemical reaction
D. to make the chemical reaction safe to observe
Solution 1
A catalyst either increases (positive catalyst) or decreases (negative catalyst) the rate of the reaction.
Question
Ammonia (NH3(g), es001-1.jpgHf = –46.19 kJ/mol) reacts with hydrogen chloride (HCl(g), es001-2.jpgHf = –92.30 kJ/mol) to form ammonium chloride (NH4Cl(s), es001-3.jpgHf = –314.4 kJ/mol) according to this equation: NH3(g) + HCl(g) es001-4.jpg NH4Cl(s) What is es001-5.jpgHrxn for this reaction? kJ
Solution 1
Answer is: enthalpy for this reaction is -175.91 kJ.
Chemical reaction: NH₃ + HCl → NH₄Cl.
ΔrH = ∑ΔfH(products of reaction) - ∑ΔfH(reactants).
ΔrH = ΔfH(NH₄Cl) - (ΔfH(NH₃) + ΔfH(HCl)).
ΔrH = -314.4 kJ/mol · 1 mol - (-46.19 kJ/mol · 1 mol + (-92.30 kJ/mol) · 1 mol).
ΔrH = -314.4 kJ + 138.49 kJ.
ΔrH = -175.91 kJ.
Chemical reaction: NH₃ + HCl → NH₄Cl.
ΔrH = ∑ΔfH(products of reaction) - ∑ΔfH(reactants).
ΔrH = ΔfH(NH₄Cl) - (ΔfH(NH₃) + ΔfH(HCl)).
ΔrH = -314.4 kJ/mol · 1 mol - (-46.19 kJ/mol · 1 mol + (-92.30 kJ/mol) · 1 mol).
ΔrH = -314.4 kJ + 138.49 kJ.
ΔrH = -175.91 kJ.
Solution 2
The value of heat of the reaction for the given chemical reaction is equal to -175.91 kJ.
How do we calculate the change in enthalpy of the reaction?
Change in enthalpy of the reaction is calculated by substracting the total sum of enthalpies of reacatnts from the the total sum of the enthalpies of products.
Given chemical reaction is:
NH₃(g) + HCl(g) → NH₄Cl(s)
According to the equation total enthalpy of the reaction calculated as:
ΔHrxn = ΔfH(NH₄Cl) - [(ΔfH(NH₃) + ΔfH(HCl)]
On putting values from the question to the equation, we get
ΔHrxn = -314.4 kJ/mol - [-46.19 kJ/mol + (-92.30 kJ/mol)]
ΔHrxn = -314.4 kJ + 138.49 kJ.
ΔHrxn = -175.91 kJ
Hence the heat of the reaction is -175.91 kJ.
To know more about total heat of reaction, visit the below link:
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