Solid lithium nitrade decomposes with heat to give lithium nitrade and oxygen gas. if 1.55 g of lithium nitrat

solid lithium nitrade decomposes with heat to give lithium nitrade and oxygen gas. if 1.55 g of lithium nitrate is decomposed, what is the milliliter volium of oxygen gas released at STP?

Answer:
2LiNO3 + heat--> 2LiNO2 + O2

1.55g LiNO3/ 68.94 g LiNO3 per mole=0.2248 mole LiNO3

.2248 mol LiNO3 forms .1124 mole O2 (two moles of LiNO3 reacts to form 1 mole O2)

at STP

#moles= Volumeof gas / 22.4 L/ mole gas

thus

#moles * 22.4 L/mole gas = volume of gas

.1124*22.4=2.51776 L==> 2517.76 ml gas

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