1. Given the following unbalanced equation: Al(SO3)3 + NaOH
---> Na2SO3 + Al(OH)3

If you start with 389.4 grams of Al(SO3)3 and 200 grams of NaOH ?

Identify the limiting reactant:

Identify the reactant in excess and by how much:

How many grams of Na2SO, will be produced:

If you perform the same reaction but get 212.4 grams of Na2SO3, what is
your percent yield?

Al2(SO3)3 + 6NaOH = 3Na2SO3 + 2Al(OH)3

(a) moles Al2(SO3)3 = 10.0 g x 1 mol/294 g = 0.0340 moles Al2(SO3)3

moles NaOH = 10.0 g x 1 mol/40 g = 0.25 moles NaOH

Divide each by the coefficient in the balanced equation:

0.0340 Al2(SO3)3 /1 = 0.0340 moles

0.25 NaOH/6 = 0.0417 moles

Al2(SO3)3 is limiting as it is in the least supply

(b) 0.0340 moles Al2(SO3)3 x 2 moles Al(OH)3/mole Al2(SO3)3 = 0.0680 moles Al(OH)3 produced

(c) 0.0340 moles Al2(SO3)3 x 3 moles Na2SO3/mole Al2(SO3)3 = 0.102 moles Na2SO3

0.102 moles Na2SO3 x 126 g/mol = 12.9 g Na2SO3 produced

(d) Excess reagent is NaOH

moles NaOH used = 0.0340 moles Al2(SO3)3 x 6 moles NaOH/mole Na2(SO3)3 = 0.204 moles used

moles NaOH remaining = 0.25 - 0.204 = 0.046 moles NaOH remaining

grams NaOH remaining = 0.046 moles NaOH x 40 g/mole = 1.84 g NaOH left over

mechanics (greek   is that area of science concerned with the behaviour of physical bodies when subjected to forces or displacements, and the subsequent effects of the bodies on their environment. the scientific discipline has its origins in ancient greece with the writings of aristotle and archimedes[1][2][3] (see history of classical mechanics and timeline of classical mechanics). during the early modern period, scientists such as galileo, kepler, and newton laid the foundation for what is now known as classical mechanics. it is a branch of classical physics that deals with particles that are either at rest or are moving with velocities significantly less than the speed of light. it can also be defined as a branch of science which deals with the motion of and forces on objects. the field is yet less widely understood in terms of quantum theory.

explanation:

"standard molar enthalpy:

h2o ( liquid water ) : - 285.8 kj/mole

h2o ( water vapor ) : - 241.8 kj / mole.

δhrxn = - 241. 8 - ( - 285.8 )= -241.8 + 285.8 = 44.0

answer: 44.0 kj"