4 Fe + 3 O2 --> 2 Fe2O3
How many grams of Fe will react with 40.0 grams of O2?

93.33 g of Fe

Explanation:

The balanced equation for the reaction is given below:

4Fe + 3O₂ —> 2Fe₂O₃

Next, we shall determine the masses of Fe and O₂ that reacted from the balanced equation. This can be obtained as follow:

Molar mass of Fe = 56 g/mol

Mass of Fe from the balanced equation = 4 × 56 = 224 g

Molar mass of O₂ = 2 × 16 = 32 g/mol

Mass of O₂ from the balanced equation = 3 × 32 = 96 g

SUMMARY:

From the balanced equation above,

224 g of Fe reacted with 96 g of O₂.

Finally, we shall determine the mass of Fe required to react with 40 g of O₂. This can be obtained as follow:

From the balanced equation above,

224 g of Fe reacted with 96 g of O₂.

Therefore, Xg of Fe will react with 40 g of O₂ i.e

Xg of Fe = (224 × 40)/96

Xg of Fe = 93.33 g

Therefore, 93.33 g of Fe is required to react with 40 g of O₂.

water at room temperature is flowing through a pipe with pressure taps located 10 ft apart along one section of the pipe, which are connected to a mercury (s=13.6) manometer.

answer is: nuclear fusion produces elements that are heavier than helium.

nuclei combine to form a heavier nucleus, releasing energy.  

for example, fusion of hydrogen atoms at the sun.  

this fusion nuclear reaction is in the inner core of the sun and hydrogen (atomic number 1) is conveted into helium (atomic number 2).  

the inner core is the central region of the sun and all solar energy is produced in the core by nuclear fusion.