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How many moles of AgCI will be formed if 10 g each of KCI and NaCI react with excess of silver nitrate? |
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Answer» `underset("one mole")(KCl) + AgNO_(3) to underset("one mole")(AgCl) darr + KNO_(3)` `underset("one mole")(NaCl) + AgNO_(3) to underset("one mole")(AgCl) darr + NaNO_(3)` The gram molecular mass of KCl `=39.01 + 35.45 = 74.55 g` `therefore` Number of moles of KCl in 10 g of it `=10/(74.55) = 0.1341` SINCE, one mole of KCI forms one mole of AGCI, the moles of AgCI formed by 0.1341 moles of KCI = 0.1341. Similarly, the number of moles of NaCI in 10 g of it `=10/(22.99 + 35.45) = 0.1711` and the number of moles of AgCI formed by 0.1711 moles of NaCI = 0.1711`therefore` The total moles of AgCI formed =0.1341 + 0.1711 =0.3052 |
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