Solve this:

Qb) i) Calculate the mass of residue obtained by heating 2.7 g. Ag2CO1 ?      [ Ag = l08  C= 12  O = 16]

       ii) How many moles of water are in 720 g of it ?

      iii) Calculate the number of atoms in 8 g of helium gas at STP.

      iv) KCIO3 on heating decomposes to give KCl and oxygen. What is the volume of O2 at stp liberated by 0.1 mole of KClO3.



 Silver carbonate decomposes as:

2Ag2CO3 (s) → 4Ag (s) + 2CO2 (g) + O2 (g)

molar mass ofAg2CO3 =  276 g

molar mass ofAg = 108g

According to the equation 2 mol ofAg2COgives 4 moles of Ag


552 g of AgNO3 gives 432 g of Ag (residue)


2.76 g of AgNO3 gives   g of Ag (residue)


Kindly post remaining query in next thread.

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B) i) i dont know
     ii) ​39.966073244495995 will be the number of moles in 720 grams of water



V=56.0 L


To solve this problem, we can first convert the given mass of He to moles, using its molar mass (4.00gmol, as seen from a periodic table):

10.0g He(1lmol He4.00g He)=2.50 mol He

At standard temperature and pressure conditions (1 bar and 0oC), one mole of an ideal gas has a volume of 22.41 L.

Therefore, using dimensional analysis again,

2.50mol He(22.41lL1mol He)=56.0 L

Thus, 10.0 grams of helium occupies a volume of roughly 56.0 liters.
this is for 10 you can substitute 10 with 8 and do the calculation
thank you plz thumbs up 

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