Chapter: Stoichiometry Examples Question 2: Determine the following A) How manyu
ID: 1036423 • Letter: C
Question
Chapter: Stoichiometry Examples
Question 2: Determine the following
A) How manyu grams of oxygen are required to burn 659 grams of propane?
B) An experimenter recovers 7.58 grams of sodium sulfrade from the neutralization of sodium hydroxide by sulfuric acid. How many grams of sodium hydroxide reacted?
C) Element A reacts with oxygen to produce a pure sample AO2. Find the atomic mass and identity of A if 9.6g of oxygen reacts with 16.5 g of A to produce 26.1 g of XO2
D) A mixture of 21.6 g or P and 78.4 g of CL2 reacts completely to form PCL3 and PCL5 wihch are the only products. Determine the mass of PCL3 that forms.
Thanks!
Explanation / Answer
A)
C3H8 +5 O2 ------------------ 3CO2 + 4 H2O
1mole 5 mole
molar mass of C3H8 = 44 gram/mole
molar mass of O2 = 32 gram/mole
according to equation
1 mole of C3H8 = 5 mole of O2
44 grams of C3H8 = 5x32 grams of O2
659 grams of C3H8 = ?
= 5x32x659/44 =2396.36 grams of O2
mass of O2 required = 2396.36 grams
B)
2 NaOH + H2SO4 ----------------- Na2SO4 + 2H2O
2 mole 1 mole
molar mass of NaOH = 40 gram/mole
Molar mass of Na2SO4 = 142.04 gram/mole
according to eqauition
1 mole of Na2SO4 = 2 moles of NaOH
142.04 grams of Na2So4 = 2x40 grams of NaOH
7.58 grams of Na2SO4 = ?
= 2x40x7.58/142.04 =4.269 grams
mass of NaOH reacted = 4.269 grams
D) 2 P + 3Cl2 -----------------2 PCl3
mass of P = 21.6 grams
molar mass of P = 31.0gram/mole
number of moles of P = 21.6/31.0 =0.6968 moles
mass of Cl2 = 78.4 grams
molar massof Cl2 = 71.0gram/mole
number of moles of Cl2 = 78.4/71 = 1.104 moles
according to equation
2 mole of P = 3 mole of Cl2
0.6968 moles of P = ?
= 3x0.6968/2 = 1.0452 moles of Cl2
we need 1.0452 mole of Cl2 but we have 1.104 moles of Cl2. so Cl2 is excess reagent.
hence P is limiting reagent
according to equation
2 moles of P = 2 moles of PCl3
0.6968 moles of P = ?
= 2x0.6968/2 = 0.6968 moles of PCl3
number of moles of PCl3 formed = 0.6968 moles
molar mass of PCl3 = 137.33 gram/mole
mass of 0.6968 moles of PCl3 = 0.6968x137.33 =95.69grams
mass of PCl3 formed = 95.69 grams.