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Nataly_w [17]
2 years ago
8

What mass in grams of iron metal can be formed by the reaction of 2.14 gAI metal with excess Fe₂O₃, according to the thermite re

action: 2Al(s) + Fe₂O₃ → 2Fe(s) + Al₂O₃(s)
Chemistry
1 answer:
kogti [31]2 years ago
8 0

Answer:

4.43 g Fe

Explanation:

To find the mass of iron, you need to (1) convert grams Al to moles Al (via molar mass), then (2) convert moles Al to moles Fe (via mole-to-mole ratio from reaction coefficients), and then (3) convert moles Fe to grams Fe (via molar mass). It is important to arrange the conversions/ratios in a way that allows for the cancellation of units (the desired unit should be in the numerator). The final answer should have 3 sig figs because the given value (2.14) has 3 sig figs.

Molar Mass (Al): 26.982 g/mol

2 Al(s) + Fe₂O₃ --->  2 Fe(s) + Al₂O₃(s)

Molar Mass (Fe): 55.845 g/mol

 2.14 g Al           1 mole              2 moles Fe           55.845 g
----------------  x  -----------------  x  --------------------  x  ------------------  =  4.43 g Fe
                         26.982 g           2 moles Al             1 mole

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273 K

Explanation:

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What is the volume of an object that has a mass of 5.80 g and a density of 6.35 g/mL?
umka2103 [35]

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1.09 mL

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8 0
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Determine the rate of the reaction shown directly below if the rate constant k is 1.1 x 10^–2 M^–2 s^–1, the NO concentration is
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6 0
3 years ago
The barium isotope 133ba has a half-life of 10.5 years. a sample begins with 1.1×1010 133ba atoms. how many are left after (a) 5
Deffense [45]
Given: Half-life of <span>133Ba = t1/2 = 10.5 years.

The radio-active materials obeys 1st order dissociation kinetics. Therefore we have:
k = 0.693 / t1/2 = 0.693 / 10.5 = 0.066 years-1.

Also, </span>k = \frac{2.303}{t} log \frac{Co}{Ct}
where, Co = initial concentration = <span>1.1×10^10 atoms
Ct = conc. of Ba at time t.
......................................................................................................................

Answer 1: For t = </span><span>5 years
</span>0.066 = \frac{2.303}{5} log \frac{Co}{Ct}
Therefore, log\frac{Co}{Ct} = 0.1432
Therefore, Co/Ct = 1.3908
Therefore, Ct = 7.9086 X 10^9 atoms.

Number of 133Ba atoms left after 5 years = 7.9086 X 10^9.
....................................................................................................................

Answer 2: For t = 30 years
0.066 = \frac{2.303}{30} log \frac{Co}{Ct}
Therefore, log\frac{Co}{Ct} = 0.8597
Therefore, Co/Ct = 7.2402
Therefore, Ct = 1.5193 X 10^9 atoms.

Number of 133Ba atoms left after 30 years = 1.5193 X 10^9.
........................................................................................................................

Answer 3: t = 180 years
0.066 = \frac{2.303}{180} log \frac{Co}{Ct}
Therefore, log\frac{Co}{Ct} = 5.1585
Therefore, Co/Ct = 1.44 X 10^5
Therefore, Ct = 7.6367 X10^4 atoms.

Number of 133Ba atoms left after 180 years = 7.6367 X10^4.
8 0
4 years ago
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