How many grams of the second reactant is needed to react with 20.0 grams of the first reactant in each balanced equation?

Question

a. P4 + 6Cl2 4PCl3
b. P4O10 + 6H2O 4H3PO4
c. Fe2O3 + 3CO 2Fe + 3CO2

Answer 1

Here is how you do the first one, step by step. This will work all stoichiometry problems so print it out and commit the process to memory.
a. P4 + 6Cl2 ==> 4PCl3
Step 1. Balanced the equation. You've done that but I added the reaction arrow.
Step 2. Convert what you have (in this case 20 g of the first reactant which is P4) to mole.
moles = grams/molar mass = 20.00/(4*P) = 20/(4*31) = 0.161
Step 3. Using the coefficients in the balanced equation, convert mols of what you have (P4) into moles of what you want (the second reactant or Cl2).
moles Cl2 = 0.161 moles P4 x (6 moles Cl2/4 moles P4) = 0.161 x 6/4 = 0.242
Step 4. Now convert moles of what you want to grams.
grams = mols x molar mass = 0.242 x (2(35.5) = 17.2 g Cl2.
Post your work for the others if you would like for me to check them for you. Just the answer will not do. I want to check your work AS WELL AS the answer.