Question

Calculate how many grams of the product form when 16.7 g of calcium metal completely reacts. Assume that there is more than enough of the chlorine gas? Ca(s) + Cl2(g) → CaCl2(s) please help me ...

Answer 1

You will get 46.2 g of calcium chloride.

Explanation:

You know that you will need a balanced chemical equation with masses, moles, and molar masses, etc.

Step 1. Assemble all the information in one place

`M_text(r):color(white)(ml)40.08color(white)(mmmm)110.98`
`color(white)(mmmm)"Ca" + "Cl"_2 → "CaCl"_2`
`m"/g": color(white)(m)16.7`

Step 2. Calculate the moles of `"Ca"`

`"Moles of Ca" = 16.7 color(red)(cancel(color(black)("g Ca"))) × ("1 mol Ca")/(40.08 color(red)(cancel(color(black)("g Ca")))) = "0.4167 mol Ca"`

Step 3 Calculate the moles of `"CaCl"_2`

`"Moles of CaCl"_2 = 0.4167 color(red)(cancel(color(black)("mol Ca")))× ("1 mol CaCl"_2)/(1 color(red)(cancel(color(black)("mol Ca")))) = "0.4167 mol CaCl"_2`

Step 4. Calculate the mass of `"CaCl"_2`

`"Mass of CaCl"_2 = 0.4167 color(red)(cancel(color(black)("mol CaCl"_2))) × "110.98 g CaCl"_2/(1 color(red)(cancel(color(black)("mol CaCl"_2)))) = "46.2 g CaCl"_2`