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4 years ago

What volume (in milliliters) of oxygen gas is required to react with 4.03 g of Mg at STP

Chemistry

2 answers:

Dvinal [7]4 years ago

7 0

Vol.250 before its to much pressure

alisha [4.7K]4 years ago

5 0

Answer: 3808 ml

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number $6.023\times 10^{23}$ of particles.

Standard condition of temperature (STP) is 273 K and atmospheric pressure is 1 atm respectively.

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}=\frac{4.03}{24}=0.17moles$

$2Mg+O_2\rightarrow 2MgO$

2 moles of magnesium react with= [tex[2\times 22.4=44.8L[/tex] of oxygen at STP

Thus 0.17 moles of magnesium react with= $\frac{44.8}{2}\times 0.17=3.8L=3808ml$

Thus the volume of oxygen required to react with 4.03 g of magnesium at STP is 3808 ml.

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I2(g) + Cl2(g)2ICl(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surroundings when 1.62 mol

galina1969 [7]

Answer:

The change in entropy of the surrounding is -146.11 J/K.

Explanation:

Enthalpy of formation of iodine gas = $\Delta H_f_{(I_2)}=62.438 kJ/mol$

Enthalpy of formation of chlorine gas = $\Delta H_f_{(Cl_2)}=0 kJ/mol$

Enthalpy of formation of ICl gas = $\Delta H_f_{(ICl)}=17.78 kJ/mol$

The equation used to calculate enthalpy change is of a reaction is:

$\Delta H_{rxn}=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]$

For the given chemical reaction:

$I_2(g)+Cl_2(g)\rightarrow 2ICl(g),\Delta H_{rxn}=?$

The equation for the enthalpy change of the above reaction is:

$\Delta H_{rxn}=[(2\times \Delta H_f_{(ICl)})]-[(1\times \Delta H_f_{(I_2)})+(1\times \Delta H_f_{(Cl_2)})]$

$=[2\times 17.78 kJ/mol]-[1\times 0 kJ/mol+1\times 62.436 kJ/mol]=-26.878 kJ/mol$

Enthaply change when 1.62 moles of iodine gas recast:

$\Delta H= \Delta H_{rxn}\times 1.62 mol=(-26.878 kJ/mol)\times 1.62 mol=-43.542 kJ$

Entropy of the surrounding = $\Delta S^o_{surr}=\frac{\Delta H}{T}$

$=\frac{-43.542 kJ}{298 K}=\frac{-43,542 J}{298 K}=-146.11 J/K$

1 kJ = 1000 J

The change in entropy of the surrounding is -146.11 J/K.

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