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3 years ago

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Neodymium battery + sulfuric acid ——> neodymium sulfate + iron sulfate + boron + hydrogen. I have 5g of Nd2Fe14B and enough H

2SO4, how much NdSO4 will I get?

1 answer:

iren2701 [21]3 years ago

7 0

Mass of Nd₂(SO₄)₃ : 2.653 g

<h3>Further explanation</h3>

Reaction

Nd₂Fe₁₄B + 17H₂SO₄ → Nd₂(SO₄)₃ + 14FeSO₄ + B + 17H₂

5 g Nd₂Fe₁₄B

MW = 2. 144,242 + 14. 55.845 + 10.811

MW = 288.484 + 781.83+10.811=1081.125 g/mol

mol :

$\tt mol=\dfrac{mass}{MW}\\\\mol=\dfrac{5}{1081.125}=0.0046$

mol Nd₂Fe₁₄B = mol Nd₂(SO₄)₃=0.0046

so mass of Nd₂(SO₄)₃ :

MW Nd₂(SO₄)₃ = 2.144.242+3.32,065+12.15,999

MW = 288.484+96.195+191.988=576.667 g/mol

$\tt mass=mol\times MW\\\\mass=0.0046\times 576.667\\\\mass=2.653~g$

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Calculate the solubility of hydrogen in water at an atmospheric pressure of 0.380 atm (a typical value at high altitude).

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The question is incomplete, here is the complete question:

Calculate the solubility of hydrogen in water at an atmospheric pressure of 0.380 atm (a typical value at high altitude).

Atmospheric Gas Mole Fraction kH mol/(L*atm)

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$O_2$ $2.10\times 10^{-1}$ $1.30\times 10^{-3}$

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<u>Explanation:</u>

To calculate the partial pressure of hydrogen gas, we use the equation given by Raoult's law, which is:

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$p_A$ = partial pressure of hydrogen gas = ?

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$\chi_A$ = mole fraction of hydrogen gas = $5.00\times 10^{-7}$

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$p_{\text{hydrogen gas}}=0.380\times 5.00\times 10^{-7}\\\\p_{\text{hydrogen gas}}=1.9\times 10^{-7}atm$

To calculate the molar solubility, we use the equation given by Henry's law, which is:

$C_{H_2}=K_H\times p_{H_2}$

where,

$K_H$ = Henry's constant = $7.80\times 10^{-4}mol/L.atm$

$p_{H_2}$ = partial pressure of hydrogen gas = $1.9\times 10^{-7}atm$

Putting values in above equation, we get:

$C_{H_2}=7.80\times 10^{-4}mol/L.atm\times 1.9\times 10^{-7}atm\\\\C_{CO_2}=1.48\times 10^{-10}M$

Hence, the solubility of hydrogen gas in water at given atmospheric pressure is $1.48\times 10^{-10}M$

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