2g sample containing NaCl ,NaBr and some inert impurities is dissolved in enough water and treated with excess of AgNO3 solution. A 3g of precipitate was formed. Precipitate on shaking with aqueous NaBr gain 0.76g of weight. Determine mass percentage of the original sample......
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NaCl + AgNO3 → AgCl + NaNO3 NaBr + AgNO3 → AgBr + NaNO3
NaCl + AgNO3 → AgCl + NaNO3 NaBr + AgNO3 → AgBr + NaNO3 Let the moles of AgCl formed be x and AgBr formed be y
NaCl + AgNO3 → AgCl + NaNO3 NaBr + AgNO3 → AgBr + NaNO3 Let the moles of AgCl formed be x and AgBr formed be yMass of AgCl formed = moles × molar mass = x × 143 =143x
NaCl + AgNO3 → AgCl + NaNO3 NaBr + AgNO3 → AgBr + NaNO3 Let the moles of AgCl formed be x and AgBr formed be yMass of AgCl formed = moles × molar mass = x × 143 =143x Mass of AgBr formed is = moles × molar mass = y × 188=188y
NaCl + AgNO3 → AgCl + NaNO3 NaBr + AgNO3 → AgBr + NaNO3 Let the moles of AgCl formed be x and AgBr formed be yMass of AgCl formed = moles × molar mass = x × 143 =143x Mass of AgBr formed is = moles × molar mass = y × 188=188y143x + 188y = 0.526 (2)
)The reaction of conversion of AgBr into AgCl by passing chlorine gas is
)The reaction of conversion of AgBr into AgCl by passing chlorine gas is2AgBr + Cl2 → 2AgCl + Br2
)The reaction of conversion of AgBr into AgCl by passing chlorine gas is2AgBr + Cl2 → 2AgCl + Br2On conversion, AgBr forms equimolar amount of AgCl
)The reaction of conversion of AgBr into AgCl by passing chlorine gas is2AgBr + Cl2 → 2AgCl + Br2On conversion, AgBr forms equimolar amount of AgCly moles of AgBr will form y moles of AgCl
)The reaction of conversion of AgBr into AgCl by passing chlorine gas is2AgBr + Cl2 → 2AgCl + Br2On conversion, AgBr forms equimolar amount of AgCly moles of AgBr will form y moles of AgClMass of y moles of AgCl = 143y
)The reaction of conversion of AgBr into AgCl by passing chlorine gas is2AgBr + Cl2 → 2AgCl + Br2On conversion, AgBr forms equimolar amount of AgCly moles of AgBr will form y moles of AgClMass of y moles of AgCl = 143y143x + 143y = 0.426
)The reaction of conversion of AgBr into AgCl by passing chlorine gas is2AgBr + Cl2 → 2AgCl + Br2On conversion, AgBr forms equimolar amount of AgCly moles of AgBr will form y moles of AgClMass of y moles of AgCl = 143y143x + 143y = 0.426 143(x + y) = 0.426
)The reaction of conversion of AgBr into AgCl by passing chlorine gas is2AgBr + Cl2 → 2AgCl + Br2On conversion, AgBr forms equimolar amount of AgCly moles of AgBr will form y moles of AgClMass of y moles of AgCl = 143y143x + 143y = 0.426 143(x + y) = 0.426 x + y = 0.426/143=0.002979
)The reaction of conversion of AgBr into AgCl by passing chlorine gas is2AgBr + Cl2 → 2AgCl + Br2On conversion, AgBr forms equimolar amount of AgCly moles of AgBr will form y moles of AgClMass of y moles of AgCl = 143y143x + 143y = 0.426 143(x + y) = 0.426 x + y = 0.426/143=0.002979y=0.002979-x
)The reaction of conversion of AgBr into AgCl by passing chlorine gas is2AgBr + Cl2 → 2AgCl + Br2On conversion, AgBr forms equimolar amount of AgCly moles of AgBr will form y moles of AgClMass of y moles of AgCl = 143y143x + 143y = 0.426 143(x + y) = 0.426 x + y = 0.426/143=0.002979y=0.002979-x Substituting the value of y in (1) and solving for x
)The reaction of conversion of AgBr into AgCl by passing chlorine gas is2AgBr + Cl2 → 2AgCl + Br2On conversion, AgBr forms equimolar amount of AgCly moles of AgBr will form y moles of AgClMass of y moles of AgCl = 143y143x + 143y = 0.426 143(x + y) = 0.426 x + y = 0.426/143=0.002979y=0.002979-x Substituting the value of y in (1) and solving for x143x + 188(0.002979-x)=0.526
)The reaction of conversion of AgBr into AgCl by passing chlorine gas is2AgBr + Cl2 → 2AgCl + Br2On conversion, AgBr forms equimolar amount of AgCly moles of AgBr will form y moles of AgClMass of y moles of AgCl = 143y143x + 143y = 0.426 143(x + y) = 0.426 x + y = 0.426/143=0.002979y=0.002979-x Substituting the value of y in (1) and solving for x143x + 188(0.002979-x)=0.5260.56-0.526 = 188x-143x
)The reaction of conversion of AgBr into AgCl by passing chlorine gas is2AgBr + Cl2 → 2AgCl + Br2On conversion, AgBr forms equimolar amount of AgCly moles of AgBr will form y moles of AgClMass of y moles of AgCl = 143y143x + 143y = 0.426 143(x + y) = 0.426 x + y = 0.426/143=0.002979y=0.002979-x Substituting the value of y in (1) and solving for x143x + 188(0.002979-x)=0.5260.56-0.526 = 188x-143x45x = 0.034 and so x = 0.0008 These are the moles of AgCl present.
)The reaction of conversion of AgBr into AgCl by passing chlorine gas is2AgBr + Cl2 → 2AgCl + Br2On conversion, AgBr forms equimolar amount of AgCly moles of AgBr will form y moles of AgClMass of y moles of AgCl = 143y143x + 143y = 0.426 143(x + y) = 0.426 x + y = 0.426/143=0.002979y=0.002979-x Substituting the value of y in (1) and solving for x143x + 188(0.002979-x)=0.5260.56-0.526 = 188x-143x45x = 0.034 and so x = 0.0008 These are the moles of AgCl present.y=0.002979-0.000757 = 0.0022 (moles of AgBr)
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