Chemical Stoichiometry: Calculating Moles of Tin Required

How many moles of Sn are required to react with 40 g of HF?

This is a grams to moles stoichiometry problem. Let's give it a try! :)

Answer:

Basically 1 mole of Sn is required.

The chemical equation given shows the reaction between tin (Sn) and hydrogen fluoride (HF) as follows:

Sn + 2HF → SnF2 + H2

The molar mass of HF is 20.01 g/mol.

For this stoichiometry problem, we start by converting the given mass of HF (40 g) to moles of Sn.

The first step is to convert the 40 g of HF to moles of HF using the molar mass of HF:

40g HF × (1 mol HF / 20.01g HF) = 1.999 moles of HF

Next, according to the balanced chemical equation, 1 mole of Sn reacts with 2 moles of HF. Therefore, we can set up a proportion:

1.999 moles HF × (1 mol Sn / 2 mol HF) = 0.999 moles Sn

Therefore, 0.999 moles of tin (Sn) are required to react with 40 grams of hydrogen fluoride (HF).

So, the answer to the question is basically 1 mole of Sn.