Consider the reaction 4Fe(s) + 3O2(g) --> 2Fe2O3(s) How many grams of oxygen are needed to produce 12.5 g of Fe2O3 (iron(III) oxide)?

3.76 g of O₂ are needed to produced 12.5 g of Fe₂O₃

Explanation:

The reaction is:  4Fe (s) + 3O₂ (g)  →  2Fe₂O₃ (s)

4 moles of iron react to 3 moles of oxygen in order to produce 2 moles of iron (III) oxide.

Let's determine the moles of the produced product.

12.5 g . 1mol/ 159.69g = 0.0783 moles

If we assume Iron in excess, we work with the oxygen.

2 moles of Fe₂O₃ are produced by 3 moles of oxygen

Then, 0.0783 moles of Fe₂O₃ might be produced by (0.0783 . 3)/2

0.117 moles.

We convert the moles to mass → 0.117 mol . 32 g/1mol = 3.76 g

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g