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ikadub [295]
2 years ago
8

An important reaction that takes place in a blast furnace during the production of iron is the formation of iron metal and CO2 f

rom Fe2O3 and CO. Part A Find the mass of Fe2O3 required to form 930 kg of iron. Express your answer with the appropriate units.
Chemistry
1 answer:
ladessa [460]2 years ago
6 0

Answer: Mass of Fe_2O_3 required to form 930 kg of iron is 1328 kg

Explanation:

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}      .....(1)

For iron:

Given mass of iron = 930 kg = 930000 g  (1kg=1000g)

Molar mass of iron = 56 g/mol

Putting values in equation 1, we get:

\text{Moles of iron}=\frac{930000g}{56g/mol}=16607mol

The chemical equation for the  production of iron  follows:

Fe_2O_3+3CO\rightarrow 2Fe+3CO_2

By Stoichiometry of the reaction:

2 moles of iron are  produced by =  1 mole of Fe_2O_3

So, 16607 moles of iron will be produced by = \frac{1}{2}\times 16607=8303moles of Fe_2O_3

Now, calculating the mass of Fe_2O_3 from equation 1, we get:

Mass of Fe_2O_3 = moles\times {\text {molar mass}}=8303\times 160=1328480g=1328kg

Thus mass of Fe_2O_3 required to form 930 kg of iron is 1328 kg

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Answer: pH=12.69

Explanation:

{\text{Moles of HF}=Molarity\times {\text{Volume of solution in liters}}

{\text{Moles of HF}=0.20M\times 0.6L=0.12 moles

HF\rightarrow H^++F^-

Initial 0.12               0       0

Eqm   0.12-x           x        x

K_a=\frac{[H^+][F^-]}{[HF]}

3.5\times 10^{-4}=\frac{x^2}{0.12-x}  

(neglecting small value of x in comparison to 0.12)

x=4.2\times 10^{-5}

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NaOH\rightarrow Na^++OH^-

{\text{Molesof NaOH}}=Molarity\times {\text{Volume of solution in liters}}

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0.06 moles of NaOH will give 0.06 moles of [OH^-]

Now 4.2\times 10^{-5} moles of OH^- will be neutralized by 4.2\times 10^{-5} moles of H^+ and (0.06-4.2\times 10^{-5})=0.059 moles of OH^- will be left.

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2 years ago
What mass of calcium carbonate is produced when 250 mL of 6.0 M sodium carbonate is added to 750 mL of 1.0 M calcium fluoride
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Volume of Na2CO3 = 250 ml = 0.250 L

Molarity of Na2CO3 = 6.0 M

Volume of CaF2 = 750 ml = 0.750 L

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<u>To determine:</u>

The mass of CaCO3 produced

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