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Burka [1]
3 years ago
11

Enter your answer in the provided box. Before arc welding was developed, a displacement reaction involving aluminum and iron(III

) oxide was commonly used to produce molten iron (the thermite process). This reaction was used, for example, to connect sections of iron railroad track. Calculate the mass of molten iron produced when 1.64 kg of aluminum reacts with 20.2 mol of iron(III) oxide.
Chemistry
1 answer:
Likurg_2 [28]3 years ago
7 0

Answer:

grams of iron = 2262.4  g

Explanation:

The balanced chemical equation should be represented first.  

2Al + Fe2O3 → Al2O3 + 2Fe

find the limiting reactants by converting to moles.

covert 1.64 kg to grams = 1640 g

moles = mass/molar mass = 1640/27 = 60.7407407407  moles

60.7407407407 × 1 mol-rxn /2 = 30.37  mol-rxn

The limiting reactant is Fe2O3 Therefore it will determine the yield of Iron

2Al + Fe2O3 → Al2O3 + 2Fe

molar mass of Fe2O3 = 56 × 2 + 16 × 3 =  112 + 48 = 160 g

atomic mass of iron = 56

mass of iron in the reaction = 56 × 2 = 112 g

moles = mass/molar mass

mass = moles × molar mass

mass = 20.2 × 160 = 3232  g

if 160 g of Fe2O3  give 112 g of iron

3232 g of Fe2O3 will give ? grams of iron

grams of iron = 3232 × 112/ 160

grams of iron = 361984 /160

grams of iron = 2262.4  g

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4 years ago
Which would have more volume? A mass of mercury or the same mass of bromine. The density of mercury is 13.5g/ml. The density of
deff fn [24]

Explanation:

Given that,

The density of mercury is 13.5 g/mL

The density of Bromine is 3.12 g/cm³

It is mentioned that Mercury and bromine have the same mass. Let d₁,d₂ are the density of Mercury and Bromine. V₁ and V₂ are their volumes. So,

\text{density}=\dfrac{\text{mass}}{\text{volume}}

\text{mass}=\text{density}\times \text{volume}

Since, mass is same.

So,

d_1V_1=d_2V_2\\\\\dfrac{d_1}{d_2}=\dfrac{V_2}{V_1}\\\\\dfrac{13.5}{3.12}=\dfrac{V_2}{V_1}\\\\4.32=\dfrac{V_2}{V_1}\\\\V_2=4.32\times V_1

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3 0
3 years ago
How many significant figures are in the number 20300?​
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3 0
4 years ago
At 506 K and 2.05 atm, if 6.00 grams of H2 will reacts with excess N2 what volume in liters will be produced of NH3? (R = 0.0820
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Answer:

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Explanation:

You must convert the 6.00 grams of H₂ into number of moles, and then use the stoichiometry of the reaction to find the number of moles of NH₃, which can be converted into volume using the ideal gas equation.

<u></u>

<u>1. Number of moles of H</u><u>₂</u>

  • number of moles = mass in grams / molar mass
  • molar mass of H₂ = 2.016g/mol
  • number of moles = 6.00grams / 2.016g/mol = 2.97619mol

<u></u>

<u>2. Number of moles of NH</u><u>₃</u>

<u></u>

i) Chemical equation:

  • 3H₂(g) + N₂(g) →  2NH₃(g)

ii) Mole ratio:

  • 3 mol H₂ : 2 mol NH₃

iii) Proportion:

  • x mol  NH₃ / 2.97619mol H₂ = 3 mol NH₃ / 2 mol H₂

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<u>3. Volume of NH₃</u>

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  • V = nRT/p
  • V = 4.4642857mol × 0.08206 atm·liter/(K·mol) × 506K / 2.05atm
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Pepsi [2]

Answer:

They do not

Explanation:

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