Question

Hey everyone I need help with this thank you! I would appreciate an explanation with the answer as well!

Answer 1

we only need 11 grams of propane to produce 36 grams of water.

Step-by-step explanation:

we can determine the number of moles of water required for 36 grams. 36g/18g/mole = 2 moles of water. we need 1/4 that number of moles of propane, so moles propane = 0.5 moles. times it's molar mass of 44 g/mole, we find that we only need 11 grams of propane to produce 36 grams of water.

Answer 2

`{\huge{\bold{\fbox{\tt{\blue{Answer}}}}}}`

______________________________________

To solve this problem, we need to use stoichiometry. Stoichiometry is a mathematical method that allows us to calculate the quantities of reactants and products in a chemical reaction.

______________________________________

First, we need to balance the chemical equation:

C3H8 + 5O2 → 3CO2 + 4H2O

______________________________________

Now, we need to determine the mole ratio between propane and water. From the balanced equation, we can see that 1 mole of propane reacts with 4 moles of water:

C3H8 + 5O2 → 3CO2 + 4H2O

1 4

______________________________________

Next, we need to determine the number of moles of water produced by 36.0 g of water vapor. We can use the molar mass of water to convert grams to moles:

molar mass of water = 18.015 g/mol

moles of water = mass / molar mass

moles of water = 36.0 g / 18.015 g/mol

moles of water = 1.998 mol

______________________________________

Finally, we can use the mole ratio to calculate the number of moles of propane required to produce 1.998 moles of water:

moles of propane = moles of water / mole ratio

moles of propane = 1.998 mol / 4

moles of propane = 0.4995 mol

______________________________________

Now, we can use the molar mass of propane to convert moles to grams:

molar mass of propane = 44.097 g/mol

grams of propane = moles of propane x molar mass of propane

grams of propane = 0.4995 mol x 44.097 g/mol

grams of propane = 22.03 g

______________________________________

Therefore, 22.03 grams of propane are required to produce 36.0 g of water vapor.

______________________________________