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

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

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.

To determine the number of grams of the second reactant needed, we first need to find the molar mass of the first reactant. Then, we can compare the coefficients in the balanced equation to calculate the required amount of the second reactant.

a. P4 + 6Cl2 -> 4PCl3

- Molar mass of P4: 4*31.0 g/mol = 124.0 g/mol
- According to the balanced equation, 6 moles of Cl2 react with 1 mole of P4.
- Therefore, the molar mass of Cl2 is: 6*35.5 g/mol = 213.0 g/mol

To find the number of grams of Cl2 needed to react with 20.0 grams of P4, we can use the following proportion:

(20.0 g P4) / (124.0 g/mol P4) = (x g Cl2) / (213.0 g/mol Cl2)

Solving for x, we find:
x = (20.0 g P4) * (213.0 g/mol Cl2) / (124.0 g/mol P4)
x ≈ 34.5 g Cl2

Therefore, approximately 34.5 grams of Cl2 is needed to react with 20.0 grams of P4 in this equation.

b. P4O10 + 6H2O -> 4H3PO4

- Molar mass of P4O10: 4*31.0 g/mol + 10*16.0 g/mol = 284.0 g/mol
- According to the balanced equation, 6 moles of H2O react with 1 mole of P4O10.
- Therefore, the molar mass of H2O is: 6*18.0 g/mol = 108.0 g/mol

Using a similar proportion as before:

(20.0 g P4O10) / (284.0 g/mol P4O10) = (x g H2O) / (108.0 g/mol H2O)

Solving for x, we find:
x = (20.0 g P4O10) * (108.0 g/mol H2O) / (284.0 g/mol P4O10)
x ≈ 7.65 g H2O

Therefore, approximately 7.65 grams of H2O is needed to react with 20.0 grams of P4O10 in this equation.

c. Fe2O3 + 3CO -> 2Fe + 3CO2

- Molar mass of Fe2O3: 2*55.8 g/mol + 3*16.0 g/mol = 159.6 g/mol
- According to the balanced equation, 3 moles of CO react with 1 mole of Fe2O3.
- Therefore, the molar mass of CO is: 3*28.0 g/mol = 84.0 g/mol

Using a similar proportion as before:

(20.0 g Fe2O3) / (159.6 g/mol Fe2O3) = (x g CO) / (84.0 g/mol CO)

Solving for x, we find:
x = (20.0 g Fe2O3) * (84.0 g/mol CO) / (159.6 g/mol Fe2O3)
x ≈ 10.5 g CO

Therefore, approximately 10.5 grams of CO is needed to react with 20.0 grams of Fe2O3 in this equation.

To find out how many grams of the second reactant are needed to react with a given amount of the first reactant, we need to use the balanced equation and the molar masses of the reactants.

Step 1: Write down the balanced equation.
a. P4 + 6Cl2 -> 4PCl3
b. P4O10 + 6H2O -> 4H3PO4
c. Fe2O3 + 3CO -> 2Fe + 3CO2

Step 2: Find the molar mass of the first reactant.
For each reactant, find the molar mass by adding up the atomic masses of all the elements in the formula.
a. Molar mass of P4 = (4 x atomic mass of P) = 4 x (30.97 g/mol) = 123.88 g/mol
b. Molar mass of P4O10 = (4 x atomic mass of P) + (10 x atomic mass of O) = 4 x (30.97 g/mol) + 10 x (16.00 g/mol) = 283.88 g/mol
c. Molar mass of Fe2O3 = (2 x atomic mass of Fe) + (3 x atomic mass of O) = 2 x (55.85 g/mol) + 3 x (16.00 g/mol) = 159.69 g/mol

Step 3: Use stoichiometry to calculate the amount of the second reactant.
a. P4 + 6Cl2 -> 4PCl3
From the balanced equation, we know that 1 mole of P4 reacts with 6 moles of Cl2 to produce 4 moles of PCl3. Using this ratio, we can find the moles of Cl2 required to react with 1 mole of P4.
Moles of Cl2 = (moles of P4) x (6 moles of Cl2 / 1 mole of P4)
To find the grams of Cl2, we can use the molar mass of Cl2 and the moles of Cl2 calculated above.
Grams of Cl2 = (moles of Cl2) x (molar mass of Cl2)

b. P4O10 + 6H2O -> 4H3PO4
From the balanced equation, we know that 1 mole of P4O10 reacts with 6 moles of H2O to produce 4 moles of H3PO4. Using this ratio, we can find the moles of H2O required to react with 1 mole of P4O10.
Moles of H2O = (moles of P4O10) x (6 moles of H2O / 1 mole of P4O10)
To find the grams of H2O, we can use the molar mass of H2O and the moles of H2O calculated above.
Grams of H2O = (moles of H2O) x (molar mass of H2O)

c. Fe2O3 + 3CO -> 2Fe + 3CO2
From the balanced equation, we know that 1 mole of Fe2O3 reacts with 3 moles of CO to produce 2 moles of Fe. Using this ratio, we can find the moles of CO required to react with 1 mole of Fe2O3.
Moles of CO = (moles of Fe2O3) x (3 moles of CO / 1 mole of Fe2O3)
To find the grams of CO, we can use the molar mass of CO and the moles of CO calculated above.
Grams of CO = (moles of CO) x (molar mass of CO)

Step 4: Calculate the grams of the second reactant.
Using the formulas derived in Step 3, plug in the calculated moles and molar masses to find the grams of the second reactant.

Note: Make sure to multiply the moles calculated in Step 3 by the given amount (in grams) of the first reactant to find the grams of the second reactant needed to react completely.

You can now follow these steps to find the grams of the second reactant for each balanced equation given in your question.