All categories

2 years ago

3Fe + 4H2O ‐----> Fe3O4 + 4H2

Chemistry

1 answer:

dexar [7]2 years ago

4 0

The limiting reactant : Fe

<h3>Further explanation</h3>

Reaction

mass of H₂O = 10 g

mol H₂O (MW=18 g/mol) :

$\tt mol=\dfrac{mass}{MW}\\\\mol=\dfrac{10}{18}=0.56$

mass of Fe = 16.8 g

mol Fe(MW=56 g/mol) :

$\tt mol=\dfrac{16.8}{56}=0.3$

Limiting reactant : the smallest ratio between mol : reaction coefficient

Fe :

$\tt \dfrac{0.3}{3}=0.1$

H₂O :

$\tt \dfrac{0.56}{4}=0.14$

Fe as limiting reactant

You might be interested in

When filling a burette for a titration, adjust the burette so that______, preferably over a sink. Then,_______to add the titrant

artcher [175]

Answer:

Explanation:

When filling a burette for a titrant, adjust the burette so that the opening is near or below the eye leve preferably over the sink.

Then, use a funnel to add the titrant into the burette.

The titrant should be filled almost to the zero mark.

6 0

2 years ago

A rigid, 2.50 L bottle contains 0.458 mol He. The pressure of the gas inside the bottle is 1.83 atm. If 0.713 mol Ar is added to

melomori [17]

Answer:The change in temperature of the gas mixture is -68.44 K.

Explanation:

1. Temperature of the bottle when only helium gas was present :T

Pressure inside the container =P= 1.83 atm

Volume of the container = V = 2.50 L

Moles of helium gas = n = 0.458 moles

Using an Ideal gas equation:

$PV=nRT$

$T=\frac{PV}{nR}=\frac{1.83 atm\times 2.50 L}{0.458 mol\times 0.0820 atm L/mol K}=121.81 K$

2. Temperature of the bottle when argon gas is added into the container:T'

Pressure inside the container = P' =2.05 atm

Volume of the container = V = 2.50 L

Moles of helium gas = n' = 0.458 mol + 0.713 mol = 1.171 mol

$P'V=n'RT'$

$T'=\frac{P'V}{n'R}=\frac{2.05 atm\times 2.50 L}{1.171 mol\times 0.0820 atm L/mol K}=53.37 K$

The change in temperature will be given as:

Final temperature = Initial temperature:T' - T

53.37 k - 121.81 K = -68.44 K

The change in temperature of the gas mixture is -68.44 K.

7 0

2 years ago

Read 2 more answers

What do you think life on earth would be like if chemistry had not been put to practical use?

tatyana61 [14]

Answer:

if chemistry hadn't been put up to practical use, we wouldn't truly understand the reason why humans are humans like makes up humans (including other understandings of biology, etc) and we wouldn't be able to have the advancements we have today (vaccines, etc).

Explanation:

5 0

1 year ago

The symbol for xenon (xe) would be a part of the noble gas notation for the element

daser333 [38]

Answer:

Explanation:

To write the electronic configuration of an element using the preceding noble gas configuration, we simply use the noble gas in the previous period of that particular element. We do not use the noble gas of that period in which the element belongs to but the one in the preceding group.

A is wrong

This is because the noble gas in the preceding period is krypton.

In fact, the electronic configuration is [Kr] 4d105s25p3

C is wrong

The noble gas in the last period here is Radon. The electronic configuration of Radium in fact is [Rn] 7s2

D is wrong

The noble gas in the last period before the period of uranium is Radon also. In fact, the electronic configuration of the element is [Rn] 5f36d17s2

B is correct, the actual electronic configuration of the element Cesium is [Xe] 6s1

6 0

2 years ago

A compound that weighs 40.8 g contains 23.6 g Sb and 17.2 g F. What is the percent composition of antimony?

Rus_ich [418]

<u>Answer:</u> The percent antimony in the compound is 57.8 %

<u>Explanation:</u>

To calculate the percentage composition of antimony in compound, we use the equation:

$\%\text{ composition of antimony}=\frac{\text{Mass of chlorine}}{\text{Mass of sample}}\times 100$

Mass of compound = 40.8 g

Mass of antimony = 23.6 g

Putting values in above equation, we get:

$\%\text{ composition of antimony}=\frac{23.6g}{40.8g}\times 100=57.8\%$

Hence, the percent antimony in the compound is 57.8 %

4 0

2 years ago