最好每一步都解释清楚.这道题让我非常抓狂……When 50.08 mL of 0.1085 M lead(II) nitrate solution are combined with 50.02 mL of 0.1012 M sodium iodide solution,a precipitate was formed that weighed 1.021 g (dry solid).Calculate the
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最好每一步都解释清楚.这道题让我非常抓狂……When 50.08 mL of 0.1085 M lead(II) nitrate solution are combined with 50.02 mL of 0.1012 M sodium iodide solution,a precipitate was formed that weighed 1.021 g (dry solid).Calculate the
最好每一步都解释清楚.这道题让我非常抓狂……
When 50.08 mL of 0.1085 M lead(II) nitrate solution are combined with 50.02 mL of 0.1012 M sodium iodide solution,a precipitate was formed that weighed 1.021 g (dry solid).
Calculate the number of moles of product (maximum amount that can be formed).
Calculate the theoretical yield of the product.
最好每一步都解释清楚.这道题让我非常抓狂……When 50.08 mL of 0.1085 M lead(II) nitrate solution are combined with 50.02 mL of 0.1012 M sodium iodide solution,a precipitate was formed that weighed 1.021 g (dry solid).Calculate the
大概翻译如下
When 50.08 mL of 0.1085 M lead(II) nitrate solution are combined with 50.02 mL of 0.1012 M sodium iodide solution,
当 50.08ml , 0.1085 mol/L 的Pb(NO3)2溶液与 50.02ml , 0.1012 mol/L 的 NaI溶液混合,
a precipitate was formed that weighed 1.021 g (dry solid).
生成1.021g 的干燥固体沉淀.
Calculate the number of moles of product (maximum amount that can be formed).
计算沉淀的分子数（沉淀能够生成的最大量【我注：就是本题产生的质量】）
Calculate the theoretical yield of the product.
计算沉淀的理论产量.
解答如下：
Pb2+和I- 可生成PbI2（2为下标,你应该能懂的） 沉淀,故而本题所说的precipitate （沉淀）指的就是PbI2.
于是第一问,已知碘化亚铅的质量是1.021g,于是其分子数为物质的量乘以阿伏加得罗常数：
N=(m/M)*NA=（1.021/(207+127*2)）*6.02*10^23=1.33*10^21（个）
第二问,沉淀理论产量,先估计谁过量,
算一算,Pb2+物质的量是：50.08*0.1085*10^(-3)=5.43368*10^(-3)mol
2I-物质的量是：50.02*0.1012/2*10^(-3)=2.531*10(-3)mol
明显,Pb2+过量,理论产量以I-计,
理论生成的沉淀的物质的量为 2.531*10^(-3)mol,故而理论产量为
2.531*10^(-3)mol*(207+127*2)g/mol=1.167g.
只是题目是英文罢了,其实内容要求不难.
最牛逼的化学题！！！
你是英文看不懂吧，这个题不算难啊
沉淀应该是碘化铅。
第一问：沉淀质量/碘化铅的相对质量*阿伏伽德罗常数
第二问：因为溶液浓度很大，并且碘化铅又是难溶物，不用考虑溶度积的问题。直接算出Pd(II), I离子那个过量，求出产量即可。