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Ifon performing theexperiment, 2 . 28 g of (NH_(4))_(2) PbCl_(6) was produced, calculatethe percentage yield of (NH_(4))_(2) Pb Cl_(6) |
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Answer» Solution :`% "yield " = ("amount produced experimentally")/("amountfor 100% yield") XX 100` `= (2 . 28)/( 4 . 56) xx 100 = 50%` |
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