The mass of oxygen gas obtained by the decompostion of 231.73 g of silver oxide is:
( Molar mass of $A g_{2} O = 231.735$ $g m o l^{- 1}$ )

16 g

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

32 g

No worries! We‘ve got your back. Try BYJU‘S free classes today!

48 g

No worries! We‘ve got your back. Try BYJU‘S free classes today!

64 g

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A
16 g
The unbalanced eqauation is,
$A g_{2} O \to A g + O_{2}$
1 mol of $A g_{2} O$ contains 1 mol of O atoms;
1 mol of $O_{2}$ contains 2 mol of O atoms.
Applying POAC on O atom,
$1 \times m o l o f$ $A g_{2} O = 2 \times m o l e o f$ $O_{2}$
Moles of $O_{2}$ = $0.5 \times m o l e s o f$ $A g_{2} O$
So, we can calculate the moles of $A g_{2} O$ , convert it into the moles of $O_{2}$ and finally to grams of $O_{2}$ .
Moles of $A g_{2} O$ = $\frac{231.73}{231.73}$ = 1 mol
Moles of $O_{2}$ = 0.5 mol
Mass of $O_{2}$ = $0.5 \times 32$ = 16 g

Suggest Corrections

Similar questions

C a O

C a C O_{3}

C O_{2}

g m o l^{- 1}

C O_{2}

g m o l^{- 1}

Hydrogen and oxygen combine in a ratio of 1:8 by mass to form water. What mass of oxygen gas is required to react completely with 3g of hydrogen gas?

Join BYJU'S Learning Program