Tin (II)fluoride can be made by the following reaction: Sn(s)+ 2 HF(g)$\rightarrow$ 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 to produce 1 mole of SnF2. With 0.480 moles of Sn and 0.720 moles of HF, HF is the limiting reactant. 0.720 moles of HF can react with 0.360 moles of Sn to produce 0.360 moles of SnF2.

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

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