Answer:
71.5g
Explanation:
The reaction equation is given as:
C + O₂ → CO₂
Mass of C = 42g
Mass of O₂ = 52g
Unknown:
Mass of CO₂ produced = ?
Solution
Now to solve this problem, we have to find limiting reactant which is the one given in short supply in this reaction.
The extent of the reaction is controlled by this reactant.
Find the number of moles of the given species;
Number of moles =
Number of moles of C =
= 3.5mol
Number of moles of O₂ =
= 1.63mol
Now;
From the balanced reaction equation;
1 mole of C reacted with 1 mole of O₂
We see that C is in excess and O₂ is the limiting reactant.
1 mole of O₂ will produce 1 mole of CO₂
So; 1.63mole of O₂ will produce 1.63 mole of CO₂
Mass of CO₂ = number of moles x molar mass
Molar mass of CO₂ = 44g/mol
Mass of CO₂ = 1.63 x 44 = 71.5g
If the temperature rises in a reaction. Exothermic is if it loses heat.
Have a nice day, brainliest would be fantastic.
Answer:
For many solids dissolved in liquid water, the solubility increases with temperature. The increase in kinetic energy that comes with higher temperatures allows the solvent molecules to more effectively break apart the solute molecules that are held together by intermolecular attractions.
Explanation:
B. A gram would have a lot more molecules of propane than a mole
Answer:
21.2 gm
Explanation:
calculate the mass of butane needed to produce 64.1 g of carbon dioxide to three significant figures and appropriate units
butane is the hydrocarbon C4H10
in combustion, we react hydrocarbons with O2 to form CO2 and H2O
so
C4H10 + O2----------------> CO2 + H2O
BALANCE
2C4H10 + 1302--------> 8CO2 + 10 H2O
the molar mass of CO2 is 12 + 16X2 = 44
64.1 gm of CO2 is
64.1/44 = 1.46 MOLES OF CO2,
FOR EVERY 8 MOLES OF CO2 WE NEED 2 MOLES OF BUTANE IT IS A
8:2 OR 4:1 RATIO. THE MOLES OF C4H10 ARE 1/4 THE MOLES OF CO2
SO
THE MOLES OF C4H10 H10 ARE 1.46/4 =0.365 MOLES
THE MOLAR MASS OF BUTANE IS 58.12
0.365 MOLES OF C4H10 HAS A MASS OF 0.365 X 58.12 = 21.2 gm