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3 years ago

If 4500 g of Zr metal is reacted with 1.5 x 104 g of water what is the limiting reactant?

Chemistry

1 answer:

PIT_PIT [208]3 years ago

6 0

Answer:

Explanation:

The balance equation of the reaction is:

Zr + 2 H₂O ⇒ ZrO₂ + 2 H₂

Being:

Zr: 91 g/mole
H: 1 g/mole
O: 16 g/mole

the molar mass of the reactants participating in the reaction are:

Zr: 91 g/mole
H₂O: 2*1 g/mole + 16 g/mole= 18 g/mole

If by stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction) they react in moles:

Zr: 1 mole
H₂O: 2 moles

then by stoichiometry the following quantity of mass react:

Zr: 1 mole*91 g/mole= 91 g
H₂O: 2 moles*18 g/mole= 36 g

Now the following rule of three applies: if 91 g of Zr react with 36 g of H₂O, 4500 g of Zr with how much mass of H₂O would it react?

$mass of H_{2} O=\frac{4500 grams of Zr*36 grams of H_{2}O }{91 grams of Zr}$

mass of H₂O= 1,780 grams

But 1780 grams of H₂0 are not available, 1.5 * 10⁴(15,000) grams are available. Since it has more mass than it needs to react with 4500 grams of Zr, Zr zirconia will be the limiting reagent.

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CO2(g)+CCl4(g)⇌2COCl2(g) Calculate ΔG for this reaction at 25 ∘C under these conditions: PCO2PCCl4PCOCl2===0.140 atm0.185 atm0.7

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Answer: The $\Delta G$ for the reaction is 54.425 kJ/mol

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$CO_2(g)+CCl_4(g)\rightleftharpoons 2COCl_2(g)$

We are given:

$\Delta G^o_f_{CO_2}=-394.4kJ/mol\\\Delta G^o_f_{CCl_4}=-62.3kJ/mol\\\Delta G^o_f_{COCl_2}=-204.9kJ/mol$

To calculate $\Delta G^o_{rxn}$ for the reaction, we use the equation:

$\Delta G^o_{rxn}=\sum [n\times \Delta G_f(product)]-\sum [n\times \Delta G_f(reactant)]$

For the given equation:

$\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(COCl_2)})]-[(1\times \Delta G^o_f_{(CO_2)})+(1\times \Delta G^o_f_{(CCl_4)})]$

Putting values in above equation, we get:

$\Delta G^o_{rxn}=[(2\times (-204.9))-((1\times (-394.4))+(1\times (-62.3)))]\\\Delta G^o_{rxn}=46.9kJ=46900J$

Conversion factor used = 1 kJ = 1000 J

The expression of $K_p$ for the given reaction:

$K_p=\frac{(p_{COCl_2})^2}{p_{CO_2}\times p_{CCl_4}}$

We are given:

$p_{COCl_2}=0.735atm\\p_{CO_2}=0.140atm\\p_{CCl_4}=0.185atm$

Putting values in above equation, we get:

$K_p=\frac{(0.735)^2}{0.410\times 0.185}\\\\K_p=20.85$

To calculate the gibbs free energy of the reaction, we use the equation:

$\Delta G=\Delta G^o+RT\ln K_p$

where,

$\Delta G$ = Gibbs' free energy of the reaction = ?

$\Delta G^o$ = Standard gibbs' free energy change of the reaction = 46900 J

R = Gas constant = $8.314J/K mol$

T = Temperature = $25^oC=[25+273]K=298K$

$K_p$ = equilibrium constant in terms of partial pressure = 20.85

Putting values in above equation, we get:

$\Delta G=46900J+(8.314J/K.mol\times 298K\times \ln(20.85))\\\\\Delta G=54425.26J/mol=54.425kJ/mol$

Hence, the $\Delta G$ for the reaction is 54.425 kJ/mol

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