Problem PageQuestion Aqueous sulfuric acid will react with solid sodium hydroxide to produce aqueous sodium sulfate and liquid water . Suppose 91. g of sulfuric acid is mixed with 116. g of sodium hydroxide. Calculate the maximum mass of sodium sulfate that could be produced by the chemical reaction. Round your answer to significant digits.

maximum mass of sodium sulphate=132g

Explanation:

firstly balance the chemical reaction,

molecular weight of H2SO4=98g/mole;

Molecular weight of NaOH=40g/mole;

number of mole of H2SO4 given=91g/(98g/mole)=0.93mole;

number of mole of NaOH given=116g/(40g/mole)=2.9mole

moles are:

number of mole of H2SO4=0.93

number of mole of NaOH=2.9

From the balanced equation,

1mole of H2SO4 reacts with 2 moles of NaOH;

hence ;

0.93 mole of H2SO4 will react with 2*0.93 moles of NaOH means

moles of NaOH that reacts wih H2SO4=1.86mole but we have given 2.9 mole therefore

NaOH will be excess reagent and H2SO4 will be the limitting reagent

and mass of product depends on limitting reagent i.e mass of H2SO4.

1mole of H2SO4 produces 1 mole of Na2SO4;

0.93 mole produces 0.93 mole of Na2SO4;

maximum mass of Na2SO4 produces=mole*molecular weight;

Molecular weight of Na2SO4=142g/mole;

maximum mass of Na2SO4 produces=0.93*142=132g;

maximum mass ofNa2SO4 produces=132g;

here we have to get the change of moles on decrease the volume of the reaction in gas phase.

the   per the given options it will decrease by (d) 11-50%

we know, on decreasing the volume of the gas phase reaction the reaction moves to the product side.

the reaction:                     so₂cl₂ (g) = so₂ (g) + cl₂ (g)         total

the moles at equilibrium:   1 - α                 α               α           = 1+α

concentrations:                   1 - α               α               α           = 1 + α

(concentration = moles/volume, as volume is 1 l)

(where α is the degree of dissociation)

now the dissociation constant in terms of concentration is :

=  

or,

or,   = (at v = 1l)

the reaction:                     so₂cl₂ (g) = so₂ (g) + cl₂ (g)         total

the moles at equilibrium:   1 - α                 α               α           = 1+α

concentrations:               (1 - α)/0.5           α/0.5         α/0.5       = (1 + α)/0.5

thus, the dissociation constant in terms of concentration is :

or,  

=  

so, the change of 1 : 1/0.5

or, 1: 2

thus it will decrease by 50%.

or, as per the given options it will decrease by 11-50%

fe 55.85 g/mol and h2o 18.02 g/mol.