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3 years ago

A balanced chemical reaction obeys the law of...

Chemistry

2 answers:

AveGali [126]3 years ago

7 0

The correct answer is C.Conservation of matter.

Anit [1.1K]3 years ago

4 0

Answer: (C) conservation of matter

Solution: Law of conservation of matter or mass states that' total mass of the reactants should always be equal to the total mass of the product that is the total mass is remained conserved in a chemical reaction.

A balanced chemical equation always follow this law.

For example:

$2H_2(g)+O_2(g)\rightarrow 2H_2O(l)$

Mass of hydrogen = 1 g/mol

Mass of Oxygen = 16 g/mol

Total mass on the reactants = 2(2×1)+(2×16)= 36g/mol

Total mass on the product side = 2[(2×1) +16] = 36 g/mol

As,

Mass on reactant side = Mass on the product side

Therefore, a balanced chemical reaction follows Law of Conservation of mass.

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3 years ago

If you pump 40.088g of octane (C8H18) into your gas tank, how mant molecules of octane are you pumping

Sonbull [250]

Answer:

2.11 × 10²³ molecules of octane

Explanation:

Given data:

Mass of octane = 40.088 g

Molecules of octane = ?

Solution:

The given problem will solve by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.

The number 6.022 × 10²³ is called Avogadro number.

For example,

18 g of water = 1 mole = 6.022 × 10²³ molecules of water

1.008 g of hydrogen = 1 mole = 6.022 × 10²³ atoms of hydrogen

Number of moles of octane:

Number of moles = mass/ molar mass

Number of moles = 40.088 g / 114.23 g/mol

Number of moles = 0.351 mol

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3 years ago

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Caluclate the number of grams of oxygen that must react

bonufazy [111]

Answer : The number of grams of oxygen react must be, 170.4 grams.

Solution : Given,

Mass of $C_3H_8$ = 46.85 g

Molar mass of $C_3H_8$ = 44 g/mole

Molar mass of $O_2$ = 32 g/mole

First we have to calculate the moles of $C_3H_8$ .

$\text{ Moles of }C_3H_8=\frac{\text{ Mass of }C_3H_8}{\text{ Molar mass of }C_3H_8}=\frac{46.85g}{44g/mole}=1.065moles$

Now we have to calculate the moles of $O_2$

The balanced chemical reaction is,

$C_3H_8+5O_2\rightarrow 3CO_2+4H_2O$

From the balanced reaction we conclude that

As, 1 mole of $C_3H_8$ react with 5 mole of $O_2$

So, 1.065 moles of $C_3H_8$ react with $1.065\times 5=5.325$ moles of $O_2$

Now we have to calculate the mass of $O_2$

$\text{ Mass of }O_2=\text{ Moles of }O_2\times \text{ Molar mass of }O_2$

$\text{ Mass of }O_2=(5.325moles)\times (32g/mole)=170.4g$

Therefore, the number of grams of oxygen react must be, 170.4 grams.

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3 years ago