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3 years ago

What is the empirical formula of a compound composed of 30.5 g potassium (k) and 6.24 g oxygen (o)?

1 answer:

5 0

K5O2

convert grams to moles, divide both by the smallest mole mass, multiply that until hole.

30.5 g K ÷ 39.10 = .78 mol
6.24 g O ÷ 16 = .39 mol
.78 mol ÷ .39 mol = 2.5
.39 mol ÷ .39 mol = 1
2.5 x 2 = 5
1 x 2 = 2
K5O2

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Liquid hexane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 6.9 g of hexane is mi

MrRa [10]

Answer:

There, there are no leftover for C6H14 instead, an additional mass of 4g of C6H14 is needed to completely react with 38.4g of O2.

Explanation:

Step 1:

We'll begin by writing the balanced equation for the reaction. This is shown below:

2C6H14 + 19O2 —> 12CO2 + 14H2O

Step 2:

Let us calculate the masses of C6H14 and O2 that reacted from the balanced equation. This is illustrated below:

2C6H14 + 19O2 —> 12CO2 + 14H2O

Molar Mass of C6H14 = (12x6) + (14x1) = 72 + 14 = 86g/mol

Mass of C6H14 from the balanced equation = 2 x 86 = 172g

Molar Mass of O2 = 16x2 =32g/mol

Mass of O2 from the balanced equation = 19 x 32 = 608g

From the balanced equation above, 172g of C6H14 reacted with 608g of O2.

Step 3.

Now, let us determine the mass of C6H14 that will react with 38.4 g of oxygen. This is illustrated below:

From the balanced equation above, 172g of C6H14 reacted with 608g of O2.

Therefore, Xg of C6H14 will react with 38.4g of O2 i.e

Xg of C6H14 = (172 x 38.4) /608

Xg of C6H14 = 10.9g

From the calculations made above, we can see clearly that the mass of C6H14 is limited as the reaction requires 10.9g of C6H14 and only 6.9g was given. There, there are no leftover for C6H14 instead, an additional mass ( 10.9 - 6.9 = 4g) of 4g of C6H14 is needed to completely react with 38.4g of O2.

5 0

2 years ago

Small quantities of oxygen can be prepared in the laboratory by heating potassium chlorate, KClO 3 ( s ) . The equation for the

Pavlova-9 [17]

Answer:

Mass of $O_2$ produced = 32 g

Explanation:

Calculation of the moles of $KClO_3$ as:-

Mass = 82.4 g

Molar mass of $KClO_3$ = 122.55 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{82.4\ g}{122.55\ g/mol}$

$Moles= 0.67237\ mol$

From the reaction shown below:-

$2KClO_3\rightarrow 2KCl+3O_2$

2 moles of potassium chlorate on reaction forms 3 moles of oxygen gas

So,

0.67237 moles of potassium chlorate on reaction forms $\frac{3}{2}\times 0.67237$ moles of oxygen gas

Moles of oxygen gas = 1 mole

Molar mass of oxygen gas = 32 g/mol

<u>Mass of $O_2$ produced = 32 g</u>

6 0

3 years ago

Fe+CuCI2> FeCI2+Cu balancing equations

Volgvan

Explanation:

To balance a chemical equation implies that we are conserving the number and mass of the reacting species.

Chemical equations must obey the law of conservation of mass "in a chemical reaction, matter is neither created nor destroyed".

This suggests that mass is conserved.

The number of atoms on the reactant side must be equal to that on the product end.

Before proceeding, check the species to see if they have equal number of atoms on both sides:

Fe + CuCI₂ → FeCI₂ + Cu

Elements reactants products

Fe 1 1

Cu 1 1

Cl 2 2

We can see that the equation is already balanced

learn more:

balanced equation brainly.com/question/2612756

#learnwithbrainly

3 0

2 years ago

Ethanol, C2H5OH, will combust in air according to the equation below. C2H5OH(l) + 3 O2(g) → 2 CO2(g) + 3 H2O(g) ΔHo = –1270 kJ/m

sergey [27]

Answer:

O2(g) was Reduced

Explanation:

It is necessary at this point to restate the working definitions of oxidation and reduction.

Oxidation refers to increase in oxidation number. A chemical specie is oxidized in a chemical reaction if there is a positive increase in its oxidation number from left to right in the reaction.

A chemical specie is said to be reduced when there is a decrease in its oxidation number from left to right in a reaction. Hence reduction refers to a decrease in oxidation number.

Now let us focus on O2(g). Its oxidation number on the left hand side is zero. On the right hand side, its oxidation number decreases to -2. This shows a decrease in oxidation number. From our premises above, we can safely conclude that O2(g) was reduced in the reaction.

6 0

2 years ago

Coefficients in a chemical equation express _____ between molecules or compounds.

Pavel [41]

Coefficients in a chemical equation express RATIOS between molecules or compounds.

In chemical equations, coefficient refers to those numerical numbers that are usually written at the front of elements' chemical symbols in order to balance the equations. Coefficients tell us the number of a particular element that must react with a specific number of another element in order for certain amount of chemical products to be produced; that is, it provide information about the reaction ratio of reactants and products. For instance, look at this chemical equation: N2 + 3H2 = 2NH3

In the equation given above, the coefficient of N2 is 1, the coefficient of H2 is 3 and the coefficient of NH3 is 2. What this equation is essentially telling us is that, in order to produce two molecules of ammonia, one molecule of nitrogen must react with three molecules of hydrogen. Thus, the equation gives the ratios between the product and the reactants.

3 0

2 years ago

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