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Calculate the weight of sodium bicarbonate to be dissociated to give `0.56` L of `CO_(2)` gas. |
Answer» `{:(2NaHCO_(3) toNaCO+HO+CO),(" 2 moles"" 1 mole"):}` GMW of `NaHCO_(3)` = 84 g 1 mole of `CO_(2)` occupies 22.4 L `22 4" of " CO_(2)` is produced from` 2 xx 84" g of "NaHCO_(3)` ` 0.56 " L of "CO_(2)` is produced from `(0.56 xx 2 84)/(22.4) xx 4.2` g |
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