Mass of Iron (Fe): 55.845Mass of Bromine (Br): 79.904
You need to multiply the mass of Br by 3 because there are 3 Bromine atoms.
(79.904)(3)+ 55.845= 239.712+55.845 = 295.557 g/mol
Answer:
400°C
Explanation:
22,000 cal / (0.11 cal/g°C x 500 g) = 400°C
The combustion reaction of propane would be expressed as:
C3H8 + 5O2 = 3CO2 + 4H2O
To determine the mass of water that is produced from the given amount of propane, we use the mass of propane and the relation of the substances from the balanced reaction. We do as follows:
moles propane = 22 g C3H8 ( 1 mol / 44.1 g ) = 0.50 mol C3H8
moles H2O = 0.50 mol C3H8 ( 4 mol H2O / 1 mol C3H8) = 2 mol H2O
mass H2O = 2 mol H2O ( 18.02 g / 1 mol ) = 36.04 g H2O
Therefore, the mass of water that is produced from 22 grams of propane would be 36.04 g.
Rarely they can't with just sight. Certain tests or experiments should take place
n = m/M = 2/18 = 1/9 ~0,1 mol
