1) A 15-g sample of lithium is reacted with 15 g of fluorine to form lithium fluoride:
2Li + F2 → 2LiF. After the reaction is complete, what will be present?

2) If 45.2 g of O2 is mixed with 45.2 g of H2 and the mixture is ignited, what is the
maximum mass of water that may be produced?

3) In a mixture of helium and chlorine, occupying a volume of 8.63 L at 890.3 mmHg and
20.8oC, it is found that the partial pressure of helium is 214.6 mmHg. What is the total
mass of the sample?

1) A 15-g sample of lithium is reacted with 15 g of fluorine to form lithium fluoride:
2Li + F2 → 2LiF. After the reaction is complete, what will be present?

Calculate the limiting reactant here
15 g of Li = 15 g X 1mole / molar mass of Li = 15 X 1/ 7 = 2.14 moles
15 g of Florine = 15 g X 1mole / molar mass of F2 = 15 X1 / 38=0.40 moles
From balanced equation
i mole of F2 reacts with two moles of Li
so 0.40 moles will react with  2 X 0.40 = 0.80 moles
so number of moles of Li left unreacted  = 2.14 - 0.80 =2.06 moles
F2 is the limiting reactant so it is used up completely
Amount of Li left = 2.06 moles X molar mass of Li = 2.06 x 7 =14.42 g
1 mole of F2 gives two moles of LiF
so amount of LiF = 0.40 X 2 = 0.80 mole X molar mass of LiF

2) If 45.2 g of O2 is mixed with 45.2 g of H2 and the mixture is ignited, what is the
maximum mass of water that may be produced?
2H2 + 1O2 ----> 2H2O
Calculate moles H2 and O2
moles of O2 = 45.2 /32= 1.41 moles
45.2 g of H2 = 45.2 / 2 =22.6 moles
1.41 moles of O2 can react with 2X 1.41 moles of H2 = 2.82 moles of H2
it means H2 is in excess and O2 is the limiting reactant
1mole of O2 gives two moles H2O
1.41 moles will give ---- 1.41 X 2= 2.82 moles of H2O

3) In a mixture of helium and chlorine, occupying a volume of 8.63 L at 890.3 mmHg and
20.8oC, it is found that the partial pressure of helium is 214.6 mmHg. What is the total
mass of the sample?

P in atm = 214.6/760 =0.282 atm
T = (20.8 + 273)K

number of moles of He = P V / RT =  0.10
mass of He = 0.10 X molar mass of He

total P = partial P of He  + partial pressure of Cl2
partial pressure of Cl2 = total P - partial P of He = 675.7 /760 atm = 0.90 atm
no of moles of Cl2 = P V /RT = 0.90 X 8.63 / 0.0821 x 293.8
than find out mass of Cl2 = number of moles X molar mass