2Ca(s) + O2(g) → 2CaO(s)
Here is a procedure that will work your stoichiometry problems.
Step 1. Write and balance the equation. You've done that.
Step 2. Convert what you have to mols. mols = g/atomic mass = 2.34/40.08=? Estimated 0.06 BUT you should use the correct value.
Step 3. Using the coefficients in the balanced equation, convert mols of what you have to mols of what you want (one of the products)
0.06 mols Ca x (2 mols CaO/2 mols Ca) = approx 0.06 mols CaO
Step 4. Convert step 3 to grams. g = mols x molar mass = about 0.06 x approx 56 = ? This is the theoretical yield in grams
Often problems will ask for a percent yield. In that case just add another step and %yield = (actual yield/theoretical yield)*100 = ?
What is the theoretic yield of CaO when you have 2.34 g of calcium which reacts with excess oxygen?
Here is formula2Ca(s) + O2(g) → 2CaO(s)
1 answer