What mass in grams of 1-chloropropane(C3H7Cl) is produced if 400.g of propane react with excess chlorine gas according to the equation C3H8 plus Cl2>>C3H7Cl plus HCl?
Well, let's do some calculations!
First, we need to find the molar mass of propane (C3H8). The molar mass of carbon (C) is approximately 12 g/mol, and the molar mass of hydrogen (H) is approximately 1 g/mol. So, the molar mass of propane is (3 * 12 g/mol) + (8 * 1 g/mol) = 44 g/mol.
Since we have 400 g of propane, we can calculate the number of moles by dividing the mass by the molar mass: 400 g / 44 g/mol = 9.09 mol of propane.
Now, let's look at the balanced equation: C3H8 + Cl2 → C3H7Cl + HCl. According to the equation, for every 1 mole of propane that reacts, we get 1 mole of 1-chloropropane (C3H7Cl).
Therefore, we can conclude that we will get 9.09 mol of 1-chloropropane as well.
Now, to find the mass of 9.09 mol of 1-chloropropane, we need to multiply by its molar mass. The molar mass of 1-chloropropane (C3H7Cl) is approximately 92 g/mol.
So, the mass of 9.09 mol of 1-chloropropane is: 9.09 mol * 92 g/mol = 830.28 g.
Therefore, approximately 830.28 grams of 1-chloropropane will be produced if 400 grams of propane reacts with excess chlorine gas. Remember, math is no joke!
To determine the mass of 1-chloropropane (C3H7Cl) produced, we first need to calculate the moles of propane (C3H8) using the given mass of 400. g and the molar mass of propane.
The molar mass of propane (C3H8) can be calculated by summing the atomic masses of carbon (C) and hydrogen (H):
3(C) + 8(H) = 3(12.01 g/mol) + 8(1.01 g/mol) = 36.04 g/mol
Next, we'll use the molar ratio from the balanced chemical equation to determine the moles of 1-chloropropane (C3H7Cl) formed for every mole of propane (C3H8) reacted. From the equation:
C3H8 + Cl2 → C3H7Cl + HCl
The molar ratio between propane and 1-chloropropane is 1:1.
Now, let's calculate the moles of propane:
moles of propane = mass of propane / molar mass of propane
= 400 g / 36.04 g/mol
≈ 11.09 mol
Since the molar ratio is 1:1, the moles of 1-chloropropane produced will also be approximately 11.09 mol.
Finally, we can calculate the mass of 1-chloropropane using the moles obtained:
mass of 1-chloropropane = moles of 1-chloropropane x molar mass of 1-chloropropane
= 11.09 mol x (12.01 g/mol + 3(1.01 g/mol) + 35.45 g/mol)
= 11.09 mol x 64.53 g/mol
≈ 716.58 g
Therefore, approximately 716.58 grams of 1-chloropropane (C3H7Cl) will be produced.
To find the mass of 1-chloropropane (C3H7Cl) produced, we need to determine the stoichiometry of the reaction. The balanced equation is:
C3H8 + Cl2 → C3H7Cl + HCl
From the equation, we can see that 1 mole of propane (C3H8) reacts with 1 mole of chlorine gas (Cl2) to yield 1 mole of 1-chloropropane (C3H7Cl) and 1 mole of hydrogen chloride (HCl).
First, we need to convert 400 g of propane to moles using its molar mass. The molar mass of propane (C3H8) is calculated as follows:
(3 atoms of C × atomic mass of C) + (8 atoms of H × atomic mass of H)
= (3 × 12.01 g/mol) + (8 × 1.01 g/mol)
= 36.03 g/mol + 8.08 g/mol
= 44.11 g/mol
To calculate the number of moles of propane, we divide the mass by the molar mass:
400 g / 44.11 g/mol ≈ 9.07 mol
According to the balanced equation, the molar ratio between propane and 1-chloropropane is 1:1. Therefore, the number of moles of 1-chloropropane produced will also be approximately 9.07 mol.
Now, we need to convert moles of 1-chloropropane into grams. The molar mass of 1-chloropropane (C3H7Cl) can be calculated as follows:
(3 atoms of C × atomic mass of C) + (7 atoms of H × atomic mass of H) + (1 atom of Cl × atomic mass of Cl)
= (3 × 12.01 g/mol) + (7 × 1.01 g/mol) + (1 × 35.45 g/mol)
= 36.03 g/mol + 7.07 g/mol + 35.45 g/mol
= 78.55 g/mol
To calculate the mass of 1-chloropropane, we multiply the number of moles by the molar mass:
9.07 mol × 78.55 g/mol ≈ 713.70 g
Therefore, approximately 713.70 grams of 1-chloropropane (C3H7Cl) will be produced when 400 grams of propane (C3H8) react with excess chlorine gas (Cl2).