All categories

4 years ago

How many grams of H2O will be produced if 750 grams of Fe are produced? Fe3O4+ H2 → Fe+H2O

Chemistry

1 answer:

Thepotemich [5.8K]4 years ago

5 0

Answer:

Roughly 323g of H2O.

Explanation:

Balance your chemical equation:

1 Fe3O4 + 4H2 → 3 Fe + 4 H2O

Modulate your stoichiometry 4 steps:

$750g \times \frac{1mol}{55.8g} \times \frac{4mol}{3mol} \times \frac{18g}{1mol} = \\ = \frac{54000}{167.4} = \\ = 322.581g$

You might be interested in

What is the atomic number of pt-175 after it undergoes alpha decay

DanielleElmas [232]

Answer:

Explanation:

Alpha decay is one of the most basic forms of radioactive decay. It has an atomic mass of 4 and a proton number of two.

Undergoing an alpha decay will decrease the atomic mass by four and the atomic number by two.

In this particular question, we are particular about the atomic number. The atom has an atomic number of 78. So taking 2 off this gives 76

5 0

3 years ago

What is it I need help

dybincka [34]

Hi, i think the correct answer would be D .

3 0

3 years ago

A ____usually requires calculations that are too complicated to do by<br>hand.

liraira [26]

Answer:

computer model

Explanation:

Computer models are cheaper to set up than alternative methods that could be used to predict what will happen in a system, ex. building a prototype. Other benefits include being able to: make alterations and quickly see the outcomes.

8 0

4 years ago

If 45.0 mL of ethanol (density=0.789 g/mL) initially at 9.0 C is mixed with 45.0 mL of water (density=1.0 g/mL) initially at 28.

Klio2033 [76]

Answer : The final temperature of the mixture is $22.7^oC$

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

And as we know that,

Mass = Density × Volume

Thus, the formula becomes,

$(\rho_1\times V_1)\times c_1\times (T_f-T_1)=-(\rho_2\times V_2)\times c_2\times (T_f-T_2)$

where,

$c_1$ = specific heat of ethanol = $2.3J/g^oC$

$c_2$ = specific heat of water = $4.18J/g^oC$

$m_1$ = mass of ethanol

$m_2$ = mass of water

$\rho_1$ = density of ethanol = 0.789 g/mL

$\rho_2$ = density of water = 1.0 g/mL

$V_1$ = volume of ethanol = 45.0 mL

$V_2$ = volume of water = 45.0 mL

$T_f$ = final temperature of mixture = ?

$T_1$ = initial temperature of ethanol = $9.0^oC$

$T_2$ = initial temperature of water = $28.6^oC$

Now put all the given values in the above formula, we get

$(0.789g/mL\times 45.0mL)\times (2.3J/g^oC)\times (T_f-9.0)^oC=-(1.0g/mL\times 45.0mL)\times 4.18J/g^oC\times (T_f-28.6)^oC$

$T_f=22.7^oC$

Therefore, the final temperature of the mixture is $22.7^oC$

4 0

3 years ago

How many moles of oxygen are needed to react with 87 grams of aluminum

labwork [276]

Answer:

2.4 moles of oxygen are needed to react with 87 g of aluminium.

Explanation:

Chemical equation:

4Al(s) + 3O₂(l) → 2AlO₃(s)

Given data:

Mass of aluminium = 87 g

Moles of oxygen needed = ?

Solution:

Moles of aluminium:

Number of moles of aluminium= Mass/ molar mass

Number of moles of aluminium= 87 g/ 27 g/mol

Number of moles of aluminium= 3.2 mol

Now we will compare the moles of aluminium with oxygen.

Al : O₂

4 : 3

3.2 : 3/4×3.2 = 2.4 mol

2.4 moles of oxygen are needed to react with 87 g of aluminium.

5 0

3 years ago