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.