Answer:
mass of excessive CO = 2.55 gram
Explanation:
Fe2O3(s)+3CO(g)→2Fe(s)+3CO2(g)
moles of Fe2O3 = mass / formula mass = 22.00/(56x2+16x3)=0.1375 (mol)
moles of CO = mass / formula mass = 14.1/(16+12) = 0.503
Fe2O3 reacts completely meanwhile CO is excessive.
mass of CO reacts = 3 x nFe2O3 x M = 3 x 0.1375 x (16+12) = 11.55 gram
mass of excessive CO = initial mass - reacted mass = 14.1 - 11.55 = 2.55 gram
Let x be the mass of the ore in grams such that the equation that would allow us to answer this item would be,
55 = (x)(0.243)
The value of x is calculated as,
x = 55/0.243
x = 226.337 grams
Then, solve for amount in kilograms by dividing the calculated value by 1000. This methodology will give us an answer of 0.226337. Hence, the answer is 0.226.
The Seafood Watch is among one of the world's in the industry because of its efficient programs in order to administer the different effects of human intervention in various marine and freshwater ecosystems. In addition to that, these programs promotes speciation in its fishing practices.
Half of the land sides to the right and the other too the left.
This statement would be best characterized by the law of conservation of momentum—choice C.
