In one of his most critical studies into the nature of combustion, Lavoisier heated mercury (II) oxide and iso

In one of his most critical studies into the nature of combustion, Lavoisier heated mercury (II) oxide and isolated elemental mercury and oxygen gas. If 41.44 g mercury (II) oxide are heated in a 747.9 mL vessel and 77.6 % (by mass) decomposes, what is the final pressure of the oxygen gas at 59.03 °C? (Assume the gas occupies the entire volume.)

I'm having a lot of trouble with this one, please help

mercury (II) oxide formula is HgO,its molecular weight is (200.6+16)= 216.6gm per mole where oxygen is 16 and mercury is 200.6 approximately
total amount of oxygen gas decomposed=(41.44gm*16*0.776)/216.6=2.38 gm
at standard temperature(273k) and pressure one mole of oxygen is 32gm and is 22.4litre
therefore 2.38 gm is equal to( 22.4*2.38)/32=1.666litre=1666ml

we know p1v1=p2v2(from boyles law)
p1=(p2v2)/v1=(1 atmosphere* 1666ml)/747.9ml=2.23atmosphere,this will be the pressure of oxygen in the vessel if the temperature is assumed to be 273k(standard)


but as temperature will increase and the pressure will also increase as the volume of the vessel is fixed
useing the formula p1v1/t1=p2v2/t2,find the pressure at standard temperature of 273k

here vessel has fixed volume so that v1=v2 and t2=59.03+273=332.03
p1/t1=p2/t2
p2=(p1*t2)/t1=(2.23 atmosphere*332.03)/273=2.71 atmosphere(answer)