When 25.0 g of Mg with 25.0 g of l2, what was the limiting reactant? How many grams of the limiting reactant are left over? How many grams of the other reactant (in excess) is left over?

Limiting reactant = I₂

Mass of Mg leftover = 22.628 g

Given

Reaction between 25.0 g of Mg with 25.0 g of l₂

Required

The limiting reactant

mass of limiting and excess left over

Solution

Reaction

Mg + I₂⇒ MgI₂

mol Mg = 25 g : 24.305 g/mol = 1.029

mol I₂ = 25 g : 253.809 g/mol = 0.098

Limiting reactant = I₂

Excess reactant = Mg

The limiting reactant will be used up

Mol excess(unreacted) :

=1.029 - 0.098

=0.931

Mass I₂ left over :

= 0.931 x 24.305

= 22.628 g

no, i was listening to water by schoolboy q while and i saw this question

explenation:

water isn't wet. wetness is a description of our experience of water; what happens to us when we come into contact with water in such a way that it impinges on our state of being. we, or our possessions, 'get wet'. a less impinging sense experience of water is that it is cold or warm, while visual experience tells us that it is green or blue or muddy or fast-flowing. we learn by experience that a sensation of wetness is associated with water: 'there must be a leak/i must have sat in something.'

"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"