How much energy is evolved during the reaction of 48.7 g of al, according to the reaction below? assume that there is excess fe2o3. fe2o3(s) + 2 al(s) → al2o3(s) + 2 fe(s) δh°rxn = - 852 kj

From the reaction, we see that for every 2 moles of Al reacted, 852 kJ of energy is released. So convert the mass given to moles:

moles Al = 48.7 g / (26.98 g/mol)

moles Al = 1.805 mol

 

So the energy evolved is:

Energy = -852 kJ / 2 mol * 1.805 mol

Energy = -769 kJ

    m₁v₁ + m₂v₂ = (m₁ + m₂)vmomentum is conserved.the left-hand side of the equation represents the momentum in the system before the collision.the right-hand side represents the momentum in the system after the collisionm₁ and v₁ represents the mass and the velocity of the physics blockm₂ and v₂ represents the mass and velocity of the ball of puttyv represents the final velocity of the combined mass.the question states the two masses stick together. they combined mass has its own velocity with the body mass the sum of the two objects' massestheir combined momentum is (m₁ + m₂)vto find the magitude of (m₁ + m₂)v, we need to add the vectors on the left-hand side of the equation: m₁v₁ + m₂v₂.note the diagram. i have positioned the vectors in accordance with their directions.the magnitude of the resultant can be calculated using the pythagorean theorem since this is a right triangle situation (north and west are perpendicular)|(m₁ + m₂)v| =  √[ (m₁v₁)² + (m₂v₂)² ]|(m₁ + m₂)v| =  √[ (1.5kg·8.0m/s)² + (0.40kg·20m/s)² ]|(m₁ + m₂)v| =  14.4 kg m/sfor two significant figures (assuming that 20m/s is two), this rounds to 14 kg m/s