1. From the data below, calculate the moles of magnesium used: Linear density of
ID: 554975 • Letter: 1
Question
1. From the data below, calculate the moles of magnesium used: Linear density of Mg ribbon: 1.084 g/m (given in lab) Length used: 6.18 cm2. From the data below, calculate the moles of hydrogen gas produced: Volume gas collected: 70.4 mL Temperature of gas: 21.4 ºC Barometric pressure: 74.77 cm Hg Vapor pressure of water at 21.4 ºC : 19.231 mmHg Hint: Use the ideal gas law equation and be sure to convert to the proper units! 1. From the data below, calculate the moles of magnesium used: Linear density of Mg ribbon: 1.084 g/m (given in lab) Length used: 6.18 cm
2. From the data below, calculate the moles of hydrogen gas produced: Volume gas collected: 70.4 mL Temperature of gas: 21.4 ºC Barometric pressure: 74.77 cm Hg Vapor pressure of water at 21.4 ºC : 19.231 mmHg Hint: Use the ideal gas law equation and be sure to convert to the proper units! 1. From the data below, calculate the moles of magnesium used: Linear density of Mg ribbon: 1.084 g/m (given in lab) Length used: 6.18 cm
2. From the data below, calculate the moles of hydrogen gas produced: Volume gas collected: 70.4 mL Temperature of gas: 21.4 ºC Barometric pressure: 74.77 cm Hg Vapor pressure of water at 21.4 ºC : 19.231 mmHg Hint: Use the ideal gas law equation and be sure to convert to the proper units!
Explanation / Answer
1. linear density= 1.084 g/m
length used= 6.18 cm
so, mass of magnesium= 1.084 g/m * 6.18 cm * 1m/100cm => 0.07 g
then moles=> mass/Molecular weight= 0.07 g/ 24 g/mol=> 0.003 moles
2. Volume=> 70.4 ml=> 70.4/1000 litre=> 0.07 L
Temperature=> 21.4°C => 21.4+273=> 294.4 K
Pressure=> 74.77 cm Hg => 74.77 cm Hg * 1atm/ 76 cm Hg=> 0.98 atm
Applying ideal gas law equation,
PV = nRT
where P is pressure
V is volume of gas
n is number of moles of gas
R is gas constant 0.0821 l atm/mol K
T is temperature.
0.98 atm * 0.07 L = n * 0.0821 L atm / mol K * 294.4 K
=> n = 0.003 moles
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