How many moles of KF can be produced when 8.00 mol K react with6.00 mol F 2 ? (I

Question

How many moles of KF can be produced when 8.00 mol K react with6.00 mol F2? (I got that right) 8 molKF
How many moles of each reactantare left over? Note that at least one must disappear. (I got that right) 0 mol K remain
>>>how many mol of F2remain?


(I got that right) 8 molKF
How many moles of each reactantare left over? Note that at least one must disappear. (I got that right) 0 mol K remain
>>>how many mol of F2remain?

Explanation / Answer

2K + F2 -> 2KF Stoichiometry ratio = n(K)/n(F2) = 2/1 = 2 Actual ratio = 8/6 = 1.33 Actual ratio

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