Tin (II)fluoride can be made by the following reaction: Sn(s)+ 2 HF(g)→ SnF₂(s)+ H₂(g). What is the maximum amount of SnF₂ that can be produced when 0.480 moles of Sn are mixed with 0.720 moles of HF?

Question

Quizplus · Accepted Answer

The reaction requires 2 moles of HF for every mole of Sn. With 0.720 moles of HF, only 0.360 moles of Sn can fully react (0.720 HF / 2 = 0.360 Sn), making it the limiting reagent and thus determining the maximum amount of SnF2 that can be produced.

Tin (II)fluoride Can Be Made by the Following Reaction: Sn(s)