The amount of oxygen needed to oxidize 12.5g of propane is 45.5g.
Stoichiometry is the calculation of the quantitative relationships between the amounts of reactants and products in chemical reactions.
The balanced chemical equation for the combustion of propane is:
C3H8 + 5O2 → 3CO2 + 4H2O
To calculate the amount of oxygen needed to react with 12.5g of propane, we need to use stoichiometry and the molar mass of propane. The molar mass of propane is 44.1 g/mol.
First, we need to convert the mass of propane to moles:
12.5 g C3H8 × (1 mol C3H8 / 44.1 g C3H8) = 0.283 mol C3H8
According to the balanced equation, 5 moles of oxygen are required to react with 1 mole of propane. So, we can calculate the amount of oxygen needed as follows:
0.283 mol C3H8 × (5 mol O2 / 1 mol C3H8) = 1.415 mol O2
Finally, we can convert the moles of oxygen to grams using the molar mass of oxygen, which is 32.0 g/mol:
1.415 mol O2 × 32.0 g O2/mol = 45.1 g O2
Therefore, the answer is e) 45.5 g of oxygen is needed to oxidize 12.5g of propane (C3H8).
Learn more about Stoichiometry here:
brainly.com/question/30215297?
#SPJ1
B. use a scale and record the weight in CM.
C.use a beaker with water and record the weight in CM.
D. use a scale and record the weight in N.
Answer:
use a scale and record the weight in N.
Explanation:
B. remain in the negatively charged body.
C. move from the positively charged body to the negatively charged body.
D. remain in the positively charged body.
Answer:
A. move from the negatively charged body to the positively charged body.
Explanation:
because this is the correct answer
Answer:
Hair is an abiotic factor because it is actually not living.