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

Explanation

The balanced equation for combustion of methane is :
`CH_(4)(g)+20_(2)(g) to CO_(2)(g)+2H_(2)O(g) `
(i) 16 g of `CH_(4)` corresponds to one mole.
(ii) From the above equation , 1 mol of `CH_(4)` (g) gives 2 mol of `H_(2)O (g)`.
2 mol of water `H_(2)O` =`2xx(2+16)
=2 xx 18=36 g`
1 mol `H_(2)O`=18g.`H_(2)O` implies `(l 8g H_(2)O)/(l "mol"H_(2)O)` =1
Hence 2 mol `H_(2)Oxx(l8g H_(2)O)/(l "mol" H_(2) O)`
= 2xx 18 g `H_(2)O=36 g H_(2)O`