Calculate the amount of water (g) produced by the combustion of 16 g of methane.

Explanation

The balance equation for combustion of methane is :
`CH_(4(g))+2O_(2(g)) rarr CO_(2(g)) + 2H_(2)O_((g))`
(i) 16 g `CH_(4)` corrasponds to one mole.
(ii) From the above equation , 1 mol of `CH_(4(g))` gives 2 mole of `H_(2)O_((g))`.
2 mol of water `H_(2)O_((g)) = 2 xx (2+16) `
`= 2 xx 18 = 36 g`
1 mol `H_(2)O_((g)) = 18 g H_(2)O = (18g H_(2)O)/(1 "mol" H_(2)O)=1`
Hence 2 mol `H_(2)O_((g)) = (18 g H_(2)O)/(1 "mol" H_(2)O)= 2 xx 18 g H_(2)O`
` = 36g H_(2)O`