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3 years ago

Write the complete balanced equation for the following reaction: c7h14 o2 yields co2 h2o

Chemistry

2 answers:

Wittaler [7]3 years ago

8 0

<u>Answer:</u> The balanced chemical equation is written below.

<u>Explanation:</u>

Balanced chemical equation is defined as the equation in which total number of individual atoms on the reactant side will be equal to the total number of individual atoms on the product side. These equations follow law of conservation of mass.

The chemical equation for the combustion of heptane follows:

$2C_7H_{14}+21O_2\rightarrow 14CO_2+7H_2O$

By Stoichiometry of the reaction:

2 moles of heptane reacts with 21 moles of oxygen gas to produce 14 moles of carbon dioxide gas and 7 moles of water molecule.

Hence, the balanced chemical equation is written above.

Svetach [21]3 years ago

3 0

C7H14 + 10.5 O2 -> 7 CO2 + 7 H2O

Or, if whole numbers must be used:

2 C7H14 + 21 O2 -> 14 CO2 + 7 H2O

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3 years ago

PLZ HELP For the reaction: 2NO2(g) → N2O4(l),

Furkat [3]

Answer: $\Delta H_{rxn}=-20kJ/mol-(+66kJ/mol)$

Explanation:

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For the given reaction :

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6 0

3 years ago

For the following reaction, 22.6 grams of nitrogen monoxide are allowed to react with 4.64 grams of hydrogen gas . nitrogen mono

antiseptic1488 [7]

Answer:

- 10.5 g of N₂

- Limiting reagent: NO

- 3.13 g of H₂ remains

Explanation:

First of all we state the reaction: 2NO(g) + 2H₂(g) → 2H₂O(l) + N₂(g)

We need to find out the limiting reactant and the excess reagent

Ratio in the reactants is 2:2. Let's convert the mass to moles:

22.6 g / 30 g/mol = 0.753 moles of NO

4.64 g / 2 g/mol = 2.32 moles of H₂

Certainly the limiting reagent is the NO and the excess reactant is the hydrogen:

- For 0.753 moles of NO, we need 0.753 moles of H₂ (we have 2.32 moles)

- For 2.32 moles of H₂, we need 2.32 moles of NO (and we don't have enough NO, because we only have 0.753 moles)

As the H₂ is the excess reagent, some moles still remains after the reaction is complete → 2.32 mol - 0.753 mol = 1.567 moles

We convert the moles to mass: 1.567 mol . 2g /1mol = 3.13 g of H₂ remains

As the NO is the limiting reagent, we can work with the equation:

We propose this rule of three: 2 moles of NO can produce 1 mol of N₂

Then, 0.753 moles of NO must produce (0.753 . 1) /2 = 0.376 moles of N₂

We convert the moles to mass 0.376 mol . 28 g / 1 mol = 10.5 g

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