What volume (in milliliters) of oxygen gas is required to react with 4.03 g of Mg at STP
2 answers:
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 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:
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=
Thus the volume of oxygen required to react with 4.03 g of magnesium at STP is 3808 ml.
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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:
Density of water = 1 g/mL
Volume of water = 25.0 mL
Putting values in above equation, we get:
To calculate the heat released by the reaction, we use the equation:
where,
q = heat released
m = Total mass = [1.25 + 25] = 26.25 g
c = heat capacity of water = 4.18 J/g°C
= change in temperature =
Putting values in above equation, we get:
To calculate the number of moles, we use the equation:
Given mass of ammonium nitrate = 1.25 g
Molar mass of ammonium nitrate = 80 g/mol
Putting values in above equation, we get:
To calculate the enthalpy change of the reaction, we use the equation:
where,
q = amount of heat released = -0.428 kJ
n = number of moles = 0.0156 moles
= enthalpy change of the reaction
Putting values in above equation, we get:
Hence, the enthalpy change of the reaction is -27. kJ/mol