If 21.4 g of aluminum is reacted with 91.3 g of iron (III) oxide, the products will be aluminum oxide and iron metal. What mass of iron will be produced?

The mass of iron produced is 44.29 grams

Explanation:

Al + FeO3    

Aluminium + Iron III oxide

2Al + F₂O₃ → Al₂O₃ + 2Fe

Molar mass of Al = 26.981539 g/mol

Molar mass of F₂O₃ = 159.69 g/mol

Number of moles of Aluminium = 21.4 g/26.981539 g/mol = 0.7931 moles

Number of moles of F₂O₃ = 91.3 g/159.69 g/mol = 0.572 moles

Hence, 1 mole of Al combines with 1/2 moles of F₂O₃, 0.7931 moles of Al will combine with 0.7931 × 1/2 or 0.3966 moles of F₂O₃ to produce 0.7931 moles of iron

Molar mass of iron = 55.845 g/mol

Therefore the mass of iron produced = Number of moles × Molar mass = 0.7931 × 55.845 = 44.29 grams.

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