关于KI和K3Fe(CN)61.5gofKIisaddedtoasolutioncontaining0.6gofK3Fe(CN)6,andthefollowingreactionoccurs:2Fe[(CN)6]^3-+2I^-2→2Fe[(CN)6]^4-+I2,怎么算KI是不是多了啊?还有howmanymolesofI2formedfromtheseamountsof

Question

关于 KI 和 K3Fe(CN)6
1.5g of KI is added to a solution containing 0.6g of K3Fe(CN)6,and the following reaction occurs:2Fe[(CN)6]^3- + 2I^-2→ 2Fe[(CN)6]^4- + I2 ,怎么算KI是不是多了啊?还有how many moles of I2 formed from these amounts of reagents?（通过这些反应物,多少摩尔的碘可以形成?）

Answer 1

n (Fe[(CN)6]^3-)=0.6/329=0.0018 mol
n (I-)=1.5/166=0.009 mol
1：2反应,KI过量.最终生成0.0018/2=0.0009mol的碘单质
里面分子量是大概计算的.我估计你写错了,题目中如果写6g K3Fe(CN)6比较合理.