Answer:
solution given:
Explanation:
4Fe +3O2------------>2Fe2O3
4mole. 3mole. 2 mole
224g. 96g. 320g
we have
224 g of fe needed to react completely with 96 g of O2
500g of fe needed to react completely with
96 ×500/224=214.3g of O2
214.3g of O2 is a required answer.
4...........................
I believe it might be c I might wrong
An oxide is a molecule that has 2 Oxygen atoms in its empirical formula. Stoichiometry would be used as the ratio of Oxygen to Dioxides is 3:2. So 52.29 would be multiplied by 3/2. The answer is 78.435 Mol of CO2 are formed.
Problem One
You will use both m * c * deltaT and H = m * heat of fusion.
Givens
m = 12.4 grams
c = 0.1291
t1 = 26oC
t2 = 1204
heat of fusion (H_f) = 63.5 J/grams.
Equation
H = m * c * deltaT + m * H_f
Solution
H = 12.4 * 0.1291 * (1063 - 26) + 12.4 * 63.5
H = 1660.1 + 787.4
H = 2447.5 or 2447.47 is the exact answer. I have to leave the rounding to you. I have no idea where to round it although I suspect 2450 would be right for 3 sig digs.
Problem Two
Formula and Givens
t1 = 14.5
t2 = 50.0
E = 5680
c = 4.186
m = ??
E = m c * deltaT
Solution
5680 = m * 4.186 * (50 - 14.5)
5680 = m * 4.186 * (35.5)
5680 = m * 148.603 * m
m = 5680 / 148.603
m = 38.22 grams That isn't very much. Be very sure you are working in joules. You'd leave that many grams in the kettle after drying it thoroughly.
m = 38.2 to 3 sig digs.
