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The weight of AgCl precipitated when a solution containing 5.85 g of NaCl is added to a solution containing 3.4g of `AgNO_3` isA. 28gB. 9.25gC. 2.870gD. 58g |
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Answer» `AgNO_3+NaCl to AgCl + NaNO_3` No. of moles of `AgNO_3=(3.4)/(170)=0.02` No. of moles of NaCl `=(5.85)/(58.5)=0.1` limiting reagent = `AgNO_3` 1 mole of `AgNO_3` produces 1 mole of AgCl 0.02 mole of `AgCO_3` produce 0.02 mole of AgCl Weight of AgCl produced = `0.02xx143.5=2.870 g` |
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