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aleksley [76]
3 years ago
5

What is the minimum mass of magnetite (Fe,O2), a iron ore, from which 120. g of pure iron could be extracted? Be sure your answe

r has a unit symbol, if necessary, and is rounded to 3 significant digits.
Chemistry
1 answer:
Schach [20]3 years ago
6 0

Answer:

166 g

Explanation:

Step 1: Write the reaction for the obtaining of Fe from magnetite

Fe₃O₄ ⇒ 3 Fe + 2 O₂

Step 2: Calculate the moles corresponding to 120 g of Fe

The molar mass of Fe is 55.85 g/mol.

120 g × (1 mol/55.85 g) = 2.15 mol

Step 3: Calculate the moles of Fe₃O₄ required to produce 2.15 moles of Fe

The molar ratio of Fe₃O₄ to Fe is 1:3. The moles of Fe₃O₄ required are 1/3 × 2.15 mol = 0.717 mol

Step 4: Calculate the mass corresponding to 0.717 moles of Fe₃O₄

The molar mass of Fe₃O₄ is 231.53 g/mol.

0.717 mol × 231.53 g/mol = 166 g

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4 0
3 years ago
A gas has a volume of 1.75L at -23°C and 150.0 kPa. At what temperature would the gas occupy 1.30L at 210.0 kPa?
Nastasia [14]

Answer:

At -13 ^{0}\textrm{C} , the gas would occupy 1.30L at 210.0 kPa.

Explanation:

Let's assume the gas behaves ideally.

As amount of gas remains constant in both state therefore in accordance with combined gas law for an ideal gas-

                                          \frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}

where P_{1} and P_{2} are initial and final pressure respectively.

           V_{1}  and V_{2} are initial and final volume respectively.

           T_{1} and T_{2} are initial and final temperature in kelvin scale respectively.

Here P_{1}=150.0kPa , V_{1}=1.75L , T_{1}=(273-23)K=250K, P_{2}=210.0kPa and V_{2}=1.30L

Hence    T_{2}=\frac{P_{2}V_{2}T_{1}}{P_{1}V_{1}}

            \Rightarrow T_{2}=\frac{(210.0kPa)\times (1.30L)\times (250K)}{(150.0kPa)\times (1.75L)}

            \Rightarrow T_{2}=260K

            \Rightarrow T_{2}=(260-273)^{0}\textrm{C}=-13^{0}\textrm{C}

So at -13 ^{0}\textrm{C} , the gas would occupy 1.30L at 210.0 kPa.

5 0
3 years ago
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