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3 years ago

Find the number of moles of water that can be formed if you have 226 mol of hydrogen gas and 108 mol of oxygen gas.

1 answer:

7 0

Answer is: there are 216 moles od water.
Chemical reaction: 2H₂ + O₂ → 2H₂O.
n(H₂) = 226 mol.
n(O₂) = 108 mol; limiting reactant.
From chemical reaction: n(O₂) : n(H₂O) = 1 : 2.
n(H₂O) = 2 · n(O₂).
n(H₂O) = 2 · 108 mol.
n(H₂O) = 216 mol.
n - amount of substance.

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3 years ago

If 7.0 mol sample of a gas has a volume of 12.2 L, what would the volume be if the amount of gas was increased to 16.8 mol

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Answer:

$V_{2} = 29.28\,L$

Explanation:

Let assume that gas behaves ideally and experiments an isobaric and isothermal processes. The following relationship is applied to determined the final volume:

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$V_{2} = V_{1} \cdot \left(\frac{n_{2}}{n_{1}} \right)$

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3 years ago

Why is oxygen written with a subscript 2 in CO2?

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4 years ago

Read 2 more answers

For a particular isomer of C 8 H 18 , C8H18, the combustion reaction produces 5104.1 kJ 5104.1 kJ of heat per mole of C 8 H 18 (

vlada-n [284]

Answer:

The standard enthalpy of formation of this isomer of octane is -220.1 kJ/mol

Explanation:

Step 1: Data given

The combustion reaction of octane produces 5104.1 kJ per mol octane

Step 2: The balanced equation

C8H18(g) + 12.5 O2 ⟶ 8CO2 (g) + 9 H2O (g) ∆H°rxn = -5104.1 kJ/mol

Step 3:

∆H°rxn = ∆H°f of products minus the ∆H° of reactants

∆H°rxn = ∆H°f products - [∆H°f reactants]

-5104.1 kJ/mol = (8*∆H°fCO2 + 9*∆H°fH20) - (∆H°fC8H18 + 12.5∆H°fO2)

∆H°f C8H18 = ∆H°f 8CO2 + ∆H°f 9H2O+ 5104.1 kJ/mol

∆H°f C8H18 = 8 * (-393.5 kJ)/mol + 9 * (-241.8 kJ/mol)] + 5104.1 kJ /mol

∆H°f C8H18 = -220.1 kJ/mol

The standard enthalpy of formation of this isomer of octane is -220.1 kJ/mol

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3 years ago

8.0 mol AgNO3 reacts with 5.0 mol Zn in

Yakvenalex [24]

Taking into account the reaction stoichiometry, 8 moles of Ag can be produced from 8 moles of AgNO₃ and 5 moles of Zn.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

2 AgNO₃ + Zn → 2 Ag + Zn(NO₃)₂

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

AgNO₃: 2 moles
Zn: 1 mole
Ag: 2 moles
Zn(NO₃)₂: 1 mole

<h3>Limiting reagent</h3>

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

<h3>Limiting reagent in this case</h3>

To determine the limiting reagent, it is possible to use a simple rule of three as follows: if by stoichiometry 1 mole of Zn reacts with 2 moles of AgNO₃, 5 moles of Zn reacts with how many moles of AgNO₃?

$amount of moles of AgNO_{3}= \frac{5 moles of Znx2 moles of AgNO_{3}}{1 mole of Zn}$

<u><em>amount of moles of AgNO₃= 10 moles </em></u>

But 10 moles of AgNO₃ are not available, 8 moles are available. Since you have less moles than you need to react with 5 moles of Zn, AgNO₃ will be the limiting reagent.

<h3>Moles of Ag formed</h3>

Considering the limiting reagent, the following rule of three can be applied: if by reaction stoichiometry 2 moles of AgNO₃ form 2 moles of Ag, 8 moles of AgNO₃ form how many moles of Ag?

$amount of moles of Ag=\frac{8 moles of AgNO_{3}x2 moles of Ag }{2 moles of AgNO_{3}}$

<u><em>amount of moles of Ag= 8 moles</em></u>

Then, 8 moles of Ag can be produced from 8 moles of AgNO₃ and 5 moles of Zn.

Learn more about the reaction stoichiometry:

<u>brainly.com/question/24741074</u>

<u>brainly.com/question/24653699</u>

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2 years ago