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

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol