Answer:
There is a mass of 154 Grams of Carbon Dioxide.
Explanation:
One mole is equal to 6.02 × 10^23 particles.
This means we have 1.05 X 10^24 total particles of Ethane.
Each ethane particle contains 2 carbon atoms.
If every particle of ethane is burned, we will end up with 2.10 x 10^24 molecules of Carbon Dioxide (Particles of Methane x 2, since each Methane particle contains 2 carbon atoms)
Carbon Dioxide has a molar mass of 44.01 g/mol
So if we take our amount of Carbon Dioxide molecules and divide it by 1 mole, ((2.10 x 10^24)/(6.02 x 10^23) = 3.49) we find that we have 3.49 moles of Carbon Dioxide.
Now all we need to do is multiply our moles of carbon dioxide(3.49) by it's molar mass(44.01) while accounting for significant digits.
What you should end up with is 154 Grams of Carbon Dioxide.
Hope this helps (And more importantly I hope I didn't make any errors in my math lol)
As a side note this is all assuming that this takes place at STP conditions.
What is the question and are there answers to go with it
dim? im not so fluent in this but i did research yesterday
The size v=masse/density
v= 4*pi*R^3
R=(3*masse/(4*pi*density))^(1/3)
R=1.9695 cm
Answer:
Mass = 11.78 g of P₄
Explanation:
The balance chemical equation is as follow:
6 Sr + P4 → 2 Sr₃P₂
Step 1: Calculate moles of Sr as;
Moles = Mass / M/Mass
Moles = 50.0 g / 87.62 g/mol
Moles = 0.570 moles
Step 2: Find moles of P₄ as;
According to equation,
6 moles of Sr reacted with = 1 mole of P₄
So,
0.570 moles of Sr will react with = X moles of P₄
Solving for X,
X = 1 mol × 0.570 mol / 6 mol
X = 0.0952 mol of P₄
Step 3: Calculate mass of P₄ as,
Mass = Moles × M.Mass
Mass = 0.0952 mol × 123.89 g/mol
Mass = 11.78 g of P₄
