Answer:
35.453 good luck with your work
First, you have to find now many moles of octane are present in 191.6g of octane. To do this you need to do this you need to divide 191.6g by its molar mass (which is 114g/mol). This will give you 1.681 moles of octane. Then you need to use the fact that 2 moles of octane are us ed to make 16 moles of carbon dioxide to find how many moles of carbon dioxide 1.681mole of octane produces. To do this you need to multiply 1.681mole by 16/2 to get 13.45mol carbon dioxide. The final step is to find the number of grams presswnt in 13.446 moles of carbon dioxide. To do this you need to multiply 13.446 mole by carbon dioxides molar mass (which is 44g/mol) to get 591.6 g of carbon dioxide.
Therefore, 591.6g of carbon dioxide is produced when 191.6 grams of octane is burned.
I hope this helps. Let me know in the comments if anything is unclear.
Answer:
The pressure changes from 2.13 atm to 1.80 atm.
Explanation:
Given data:
Initial pressure = ?
Final pressure = 1.80 atm
Initial temperature = 86.0°C (86.0 + 273 = 359 K)
Final temperature = 30.0°C (30+273 =303 K)
Solution:
According to Gay-Lussac Law,
The pressure of given amount of a gas is directly proportional to its temperature at constant volume and number of moles.
Mathematical relationship:
P₁/T₁ = P₂/T₂
Now we will put the values in formula:
P₁ = P₂T₁ /T₂
P₁ = 1.80 atm × 359 K / 303 K
P₁ = 646.2 atm. K /303 K
P₁ = 2.13 atm
The pressure changes from 2.13 atm to 1.80 atm.
Oxygen gas produced : 0.7 g
<h3>Further explanation</h3>
Given
10.0 grams HgO
9.3 grams Hg
Required
Oxygen gas produced
Solution
Reaction⇒Decomposition
2HgO(s)⇒2Hg(l)+O₂(g)
Conservation of mass applies to a closed system, where the masses before and after the reaction are the same
mass of reactants = mass of products
mass HgO = mass Hg + mass O₂
10 g = 9.3 g + mass O₂
mass O₂ = 0.7 g
