I balanced
Li(s) + Cl2(g)= LiCl(s) to
2Li(s) + Cl2(g)= 2LiCl(s)
Now the question is, if you have 39.8g Li, how many moles of LiCl would you have?
I am stuck on this part. Any help would be appreciated.
Divide 39.8 g by the atomic weight of Li, 6.94 g/mole. That gives you 5.735 moles of lithium reactant.
You form the same number of moles of LiCl product.
Thank you!
To determine the number of moles of LiCl produced, we need to use the balanced equation and the molar mass of LiCl. Here's how you can do it step by step:
Step 1: Calculate the molar mass of LiCl.
The molar mass of lithium (Li) is approximately 6.94 g/mol, and the molar mass of chlorine (Cl) is approximately 35.45 g/mol. To find the molar mass of LiCl, add these two values together:
Molar Mass of LiCl = Molar Mass of Li + Molar Mass of Cl
= 6.94 g/mol + 35.45 g/mol
= 42.39 g/mol
Step 2: Convert the given mass of Li to moles of Li using its molar mass.
Given mass of Li = 39.8 g
Moles of Li = Given Mass of Li / Molar Mass of Li
= 39.8 g / 6.94 g/mol
≈ 5.73 mol
Step 3: Use the stoichiometric coefficients from the balanced equation to determine the moles of LiCl produced.
The balanced equation is: 2Li(s) + Cl2(g) → 2LiCl(s)
According to the equation, 2 moles of LiCl are produced for every 2 moles of Li. So, the moles of LiCl produced are the same as the moles of Li:
Moles of LiCl = 5.73 mol
Therefore, if you have 39.8 g of Li, you would have approximately 5.73 moles of LiCl.