A gas occupies a volume of 750 mL at 101.3 kPa. What pressure (in kPa) is needed to decrease the volume to 250mL?
1 answer:
Answer:
The final pressure of the gas comes out to be 303.9 KPa
Explanation:
Initial volume of gas = V = 750 mL
Initial pressure of gas = P = 101.3 KPa
Final volume of gas = V' = 250 mL
Assuming final pressure of the gas to be P' KPa.
Assuming temperature to be kept constant.
The final pressure can be obtained by following expression shown below
Final pressure of gas = 303.9 KPa
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Answer:
2.24 L of hydrogen gas, measured at STP, are produced.
Explanation:
Given, Moles of magnesium metal, = 0.100 mol
Moles of hydrochloric acid, = 0.500 mol
According to the reaction shown below:-
1 mole of Mg reacts with 2 moles of
0.100 mol of Mg reacts with 2*0.100 mol of
Moles of must react = 0.200 mol
Available moles of = 0.500 moles
Limiting reagent is the one which is present in small amount. Thus, is limiting reagent.
The formation of the product is governed by the limiting reagent. So,
1 mole of on reaction forms 1 mole of
0.100 mole of on reaction forms 0.100 mole of
Mole of = 0.100 mol
At STP,
Pressure = 1 atm
Temperature = 273.15 K
Volume = ?
Using ideal gas equation as:
where,
P is the pressure
V is the volume
n is the number of moles
T is the temperature
R is Gas constant having value = 0.0821 L.atm/K.mol
Applying the equation as:
1 atm × V L = 0.100 × 0.0821 L.atm/K.mol × 273.15 K
<u>⇒V = 2.24 L</u>
2.24 L of hydrogen gas, measured at STP, are produced.