Calculate the percent weight loss suffered by sodium bicarbonate on strong heating? 

Explanation

On strong heationg sodium bicarbonate decomposes and loses carbondioxide and water.
`2NaHCO_(3) to Na_(2)CO_(3)+H_(2)O+CO_(2)`
2 moles of `NaHCO_)(3)="1 mole of "CO_(2)+"1 mole of "H_(2)O`
`(2 xx 84)" grams of "NaHCO_(3)-=(44+18)` grams weight loss 100 grams of `NaHCO_(3)-=?`
The percent weight loss suffered by sodium carbonate on calcination `=(100)/(2 xx 84)xx 62=36.9%`