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When 0.500 l of a 1.00 m solution of pb no3 2 is mixed with an excess of 1.00 m solution of ki how many grams of lead II iodide are formed?
Andrewryanharrington
approx 2.4354g
Wiki User
∙ 14y ago
The reaction between lead (II) nitrate (Pb(NO3)2) and potassium iodide (KI) will form lead (II) iodide (PbI2). The balanced chemical equation is 2Pb(NO3)2 + 4KI -> 2PbI2 + 4KNO3. Given that 0.500 L of 1.00 M Pb(NO3)2 is used, this corresponds to 0.500 moles. Since the reaction ratio is 2:2 between Pb(NO3)2 and PbI2, 0.500 moles of Pb(NO3)2 will form 0.500 moles of PbI2, which is equivalent to 0.500 * 461 g = 230.5 grams of PbI2.
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