Help with highschool STOICH problem please!!

Question

How many grams of Chloroform (CHCI₃) are formed if you have 30 grams of CH₂Cl₂and a room full of Chlorine gas?

CH₂Cl₂ (g)+ Cl₂(g) = CHCL₃ (I) + HCI (g)

Answer 1

CH2Cl2 + Cl2 ----------> CHCl3 + HCl

CH2Cl2 is the limiting reactant, so it will determine the outcome.

1 mole of CH2Cl2 gives 1 mole of CHCl3.

1 mole of CH2Cl2 = 12.01 + ( 1.01 * 2 ) + ( 35.45 * 2 ) g

=12.01 +2.02 +70.9 = 84.93 g

1 mole of CHCl3 = 12.01 + 1.01 + ( 35.45 * 3 ) g

= 12.01 + 1.01 + 106.35 = 119.37 g

84.93 g ------> 119.37 g

30.0 g ------> X g

119.37 g * 30 g

X g = ------------------------ = 42.166 g

84.93 g

Answer 2

How many grams of Chloroform (CHCI₃) are formed if you have 30 grams of CH₂Cl₂and a room full of Chlorine gas?

CH₂Cl₂ (g)+ Cl₂(g) = CHCL₃ (I) + HCI (g)

**When the question gives you grams of a reactant and wants grams of a product, you will definitely be using stoichiometry. It would be helpful to first find the molar mass of the reactant and product. Molar mass can be found by using the masses on the periodic table.** Keep in mind that molar mass is always grams per mole.

CH2Cl2 = 12.01 + (1.01*2) + (35.45*2) = 84.93 g/mol

CHCl3 = 12.01 + 1.01 + (35.45*3) = 119.37 g/mol

1. Start with the given in the question (30 g CH2Cl2)
2. To cancel grams CH2Cl2, put g CH2Cl2 down and diagonal.
3. We are converting to moles CH2Cl2, so put moles CH2Cl2 on top.
4. The molar mass will ALWAYS go with grams, not moles! So we put the 84.93 with the grams on the bottom and 1 with mole on the top.
5. To switch substances, you must use a mole to mole ratio. How do you find the mole to mole ratio? **You use the coefficients from the balanced chemical equation**
6. First, we make sure the equation is balanced.
7. If it wasn't, we would need to balance it.
8. If the equation wasn't written out at all, we would need to write it out.
9. This equation is balanced all of the coefficients are 1.
10. We put 1 mol of CH2Cl2 on the bottom to cancel it out, and we put 1 mol CHCl3, which is what we want on the top.
11. We need to convert the mol of CHCl3 into grams. **This where we need molar mass**
12. To cancel mol of CHCl3, we put moles on the bottom, and grams on the top.
13. The molar mass ALWAYS goes with grams, so the 119.37 with grams on top.
14. You multiply everything on top and divide by everything on the bottom.

30 g CH2Cl2 | 1 mol CH2Cl2 | 1 mol CHCl3 | 119.37 g

------------------------------------------------------------------------------------ = 42.17 g CHCl3

| 84.93 g CH2Cl2 | 1 mol CH2Cl2 | 1 mol CHCl3