How many grams of iron are needed to combine with 24.9 g of 0 to make fe203?

Explanation:

Here's where all that equation balancing is going to come into use. Since the main object of the question is not the equation, I'm just going to balance it and use it.

4Fe + 3O2 > 2Fe2O3

Step One

Find the number of mols of O2 in 24.9 grams of O2

1 mol O2 = 2*16 = 32 grams

x mol O2 = 24.9 grams               Cross multiply

32x = 24.9 * 1                               Divide by 32

x = 24.9/32

x = 0.778 moles of O2

Step Two

Type the findings under the balanced equations parts. Solve for the number of moles of Fe

4Fe + 3O2 > 2Fe2O3

x          0.778

Step Three

Set up the proportion

4/x = 3/0.778            Cross multiply

Step Four

Solve the proportion moles of Fe

4*0.778 = 3x              

3.112   =    3x                      Divide by 3

3.112/3 = 3x/3

x = 1.037 moles of Fe

Step Five

Find the mass of Fe

1 mol Fe = 56 grams

1.037 mol Fe = x            Cross Multiply

x = 56*1.037

x = 58.1 grams

the equation that correctly represent the reaction for formation of ammonia is

n2+ 3h2 → 2nh3 (answer d)

1 mole of nitrogen gas (n2) react with 3 moles of hydrogen gas (h2) to form 2 moles of ammonia ( nh3). this is in a process known as haber process were iron is used as a catalyst and reaction take place that a higher temperature and pressure. the process is exothermic hence energy is released.