All categories

2 years ago

If you need to produce 9.75 moles of iron(III) oxide (Fe2O3), how many moles of iron will be required? *

Chemistry

1 answer:

torisob [31]2 years ago

7 0

Answer: Thus 9.75 moles of iron will be required.

Explanation:

The balanced chemical reaction is:

$2Fe+3O_2\rightarrow 2Fe_2O_3$

According to stoichiometry :

2 moles of $Fe_2O_3$ are produced by= 2 moles of $Fe$

Thus 9.75 moles of $Fe_2O_3$ are produced by= $\frac{2}{2}\times 9.75=9.75moles$ of $Fe$

Thus 9.75 moles of iron will be required.

You might be interested in

How many grams of carbonic acid were produced by the 3.00 g sample of NaHCO

stellarik [79]

I think the given is 3 g sample of NaHCO3. then if it will be reacted with an acid, it will produce H2CO3.
so the reaction NaHCO3 + HCl --> NaCl + H2CO3

mas of H2CO3 = 3 g NaHCO3 ( 1 mol NaHCO3 / 84 g ) ( 1 mol H2CO3 / 1 mol NaHCO3) ( 62.03 g / 1 mol )
mass of H2CO3 = 2.22 g H2CO3

7 0

3 years ago

Which three statements about gravity and the formation of the solar system

Delicious77 [7]

Answer:

B. Gravity held the pieces of forming planets together.

c. Gravity pulled most of the matter into the center of the solar system

D. Gravity caused the planets and Sun to have spherical shapes.

Explanation:

When a collection of grains pulled together by their gravitational forces would keep in by the gravity of a star, it would eventually became bigger to the point a planet was formed.

The sun's strong gravitational force pulled most of the matter around it to the center of the solar system.

The spherical shape of planets is a result of their gravity pulling equally from all sides, shaping it into a sphere.

7 0

2 years ago

Read 2 more answers

A compound is composed of C, H and O. A 1.621 g sample of this compound was combusted, producing 1.902 g of water and 3.095 g of

vlada-n [284]

Answer: The molecular of the compound is, $C_2H_3O$

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=3.095g$

Mass of $H_2O=1.902g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in $3.095g$ of carbon dioxide, $\frac{12}{44}\times 3.095=0.844g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in $1.902g$ of water, $\frac{2}{18}\times 1.092=0.121g$ of hydrogen will be contained.

For calculating the mass of oxygen:

Mass of oxygen in the compound = $(1.621)-[(0.844)+(0.121)]=0.656g$

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.844g}{12g/mole}=0.0703moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.121g}{1g/mole}=0.121moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.656g}{16g/mole}=0.041moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is $0.041$ moles.

For Carbon = $\frac{0.0703}{0.041}=1.71\approx 2$

For Hydrogen = $\frac{0.121}{0.041}=2.95\approx 3$

For Oxygen = $\frac{0.041}{0.041}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 2 : 3 : 1

Hence, the empirical formula for the given compound is $C_2H_3O_1=C_2H_3O$

The empirical formula weight = 2(12) + 3(1) + 1(16) = 43 gram/eq

Now we have to calculate the molecular formula of the compound.

Formula used :

$n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}$

$n=\frac{46.06}{43}=1$

Molecular formula = $(C_2H_3O_1)_n=(C_2H_3O_1)_1=C_2H_3O$

Therefore, the molecular of the compound is, $C_2H_3O$

6 0

3 years ago

This table shows the densities of a sample of substances.

Arisa [49]

When placed in a container, the heaviest (most dense) will sink to the bottom and the lightest (least dense) will rise to the top.
Therefore, Gasoline would rise to the top.

3 0

3 years ago

Can someone help?? This is really hard.

Sergeu [11.5K]

Answer:

See Explanation

Explanation:

Let us consider the first two reactions, the initial concentration of CO was held constant and the concentration of Hbn was doubled.

2.68 * 10^-3/1.34 * 10^-3 = 6.24 * 10^-4/3.12 * 10^-4

2^1 = 2^1

The rate of reaction is first order with respect to Hbn

Let us consider the third and fourth reactions. The concentration of Hbn is held constant and that of CO was tripled.

1.5 * 10^-3/5 * 10^-4 = 1.872 * 10^-3/6.24 * 10^-4

3^1 = 3^1

The reaction is also first order with respect to CO

b) The overall order of reaction is 1 + 1=2

c) The rate equation is;

Rate = k [CO] [Hbn]

d) 3.12 * 10^-4 = k [5 * 10^-4] [1.34 * 10^-3]

k = 3.12 * 10^-4 /[5 * 10^-4] [1.34 * 10^-3]

k = 3.12 * 10^-4/6.7 * 10^-7

k = 4.7 * 10^2 mmol-1 L s-1

e) The reaction occurs in one step because;

1) The rate law agrees with the experimental data.

2) The sum of the order of reaction of each specie in the rate law gives the overall order of reaction.

8 0

3 years ago