Respuesta :
The reaction equation may be written as:
2Li + 2H₂O → 2LiOH + H₂
A single mole of gas occupies 22,400 ml of gas at STP. Therefore, the moles of hydrogen gas are
75.5 / 22,400 = 0.0034 mol
According to the equation, the moles of lithium are twice that of hydrogen gas; thus,
0.0034 x 2 = 0.0068 mol
The mass of lithium is given by:
mass = moles x atomic mass
mass = 0.0068 x 6.9
mass = 0.047 grams
2Li + 2H₂O → 2LiOH + H₂
A single mole of gas occupies 22,400 ml of gas at STP. Therefore, the moles of hydrogen gas are
75.5 / 22,400 = 0.0034 mol
According to the equation, the moles of lithium are twice that of hydrogen gas; thus,
0.0034 x 2 = 0.0068 mol
The mass of lithium is given by:
mass = moles x atomic mass
mass = 0.0068 x 6.9
mass = 0.047 grams
The reaction would written as:
2Li + 2H2O = 2LiOH + H2
We are given the amount in volume of the hydrogen gas that is produced. Assuming that this gas is an ideal gas, we use the relation that in every 1 mol of gas, 22.4 L is being occupied.
75.5 mL ( 1 L / 1000 mL ) ( 1 mol / 22.4 L ) = 0.0034 mol H2 produced
0.0034 mol H2 ( 2 mol Li / 1 mol H2 ) ( 6.941 g / mol ) = 0.0472 g Li is needed
2Li + 2H2O = 2LiOH + H2
We are given the amount in volume of the hydrogen gas that is produced. Assuming that this gas is an ideal gas, we use the relation that in every 1 mol of gas, 22.4 L is being occupied.
75.5 mL ( 1 L / 1000 mL ) ( 1 mol / 22.4 L ) = 0.0034 mol H2 produced
0.0034 mol H2 ( 2 mol Li / 1 mol H2 ) ( 6.941 g / mol ) = 0.0472 g Li is needed