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Jlenok [28]
2 years ago
11

If you have 23 g of Lithium oxide, what is the volume of Oxygen that is produced? _____ Li2O2 + _____ H2O  _____ LiOH + _____ O

2
Chemistry
1 answer:
Svet_ta [14]2 years ago
8 0

Answer:

1st step : find the number of moles for the known substance

number of moles =mass/ relative molecular mass

n=23/Li2O2

n=23/(14+32)

n =23/46

n= 0,5moles

2nd step : balance the equation

2Li202 +2H20-----> 4LiOH +O2

3rd step : find the mole ratio of the known or stated using the balanced equation

Li2O2 : O2

2 : 1

4th step :find the number of moles for the asked using the mole ratio using simple proportion

Li2O2 :O2

2 :1

0,5 :???

0,5/2 ×1

= 0,25moles for (O2)

5th step : find the asked either mass , volume , concentration

number of moles = volume/molar gas volume

volume = number of moles × molar gas volume

v = 0,25moles × 28dm^3

v= 7 dm^3 of (O2)

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<u>Answer:</u> The enthalpy change of the reaction is -27. kJ/mol

<u>Explanation:</u>

To calculate the mass of water, we use the equation:

\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}

Density of water = 1 g/mL

Volume of water = 25.0 mL

Putting values in above equation, we get:

1g/mL=\frac{\text{Mass of water}}{25.0mL}\\\\\text{Mass of water}=(1g/mL\times 25.0mL)=25g

To calculate the heat released by the reaction, we use the equation:

q=mc\Delta T

where,

q = heat released

m = Total mass = [1.25 + 25] = 26.25 g

c = heat capacity of water = 4.18 J/g°C

\Delta T = change in temperature = T_2-T_1=(21.9-25.8)^oC=-3.9^oC

Putting values in above equation, we get:

q=26.25g\tiimes 4.18J/g^oC\times (-3.9^oC)=-427.9J=-0.428kJ

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

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Given mass of ammonium nitrate = 1.25 g

Molar mass of ammonium nitrate = 80 g/mol

Putting values in above equation, we get:

\text{Moles of ammonium nitrate}=\frac{1.25g}{80g/mol}=0.0156mol

To calculate the enthalpy change of the reaction, we use the equation:

\Delta H_{rxn}=\frac{q}{n}

where,

q = amount of heat released = -0.428 kJ

n = number of moles = 0.0156 moles

\Delta H_{rxn} = enthalpy change of the reaction

Putting values in above equation, we get:

\Delta H_{rxn}=\frac{-0.428kJ}{0.0156mol}=-27.44kJ/mol

Hence, the enthalpy change of the reaction is -27. kJ/mol

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