<span>the balanced chemical equation for the reaction is as follows;
C</span>₃H₈ + 5O₂ ---> 3CO₂ + 4H₂<span>O
stoichiometry of </span> C₃H₈ to O₂ is 1:5
number of moles of C₃H₈ reacted - 0.025 g / 44.1 g/mol = 0.000567 mol according to molar ratio of 1:5
number of O₂ moles required are 5 times the amount of C₃H₈ moles reacted therefore number of O₂ moles required - 0.000567 x 5 = 0.00284 mol .
mass of O₂ required - 0.00284 mol x 32.00 g/mol = 0.091 mol .
answer is 0.091 mol
Answer:
2.6 ×10^-42
Explanation:
From
∆G= -RTlnK
∆G= -237.2 KJmol-1 or -237.2×10^3 Jmol-1
R= 8.314 Jmol-1K-1
T= 25°C + 273= 298K
-237.2×10^3= 8.314 × 298 × ln K
ln K= -237.2×10^3/2477.572
K = 2.6 ×10^-42
Take the 72 g and divid it by 6 which would equal 12 g each
The answer is A.Hope this helps
Answer:
The concentration the student should write down in her lab is 2.2 mol/L
Explanation:
Atomic mass of the elements are:
Na: 22.989 u
S: 32.065 u
O: 15.999 u
Molar mass of sodium thiosulfate, Na2S2O3 = (2*22.989 + 2*32.065 + 3*15.999) g/mol = 158.105 g/mol.
Mass of Na2S2O3 taken = (19.440 - 2.2) g = 17.240 g.
For mole(s) of Na2S2O3 = (mass taken)/(molar mass)
= (17.240 g)/(158.105 g/mol) = 0.1090 mole.
Volume of the solution = 50.29 mL = (50.29 mL)*(1 L)/(1000 mL)
= 0.05029 L.
To find the molar concentration of the sodium thiosulfate solution prepared we use the formula:
= (moles of sodium thiosulfate)/(volume of solution in L)
= (0.1090 mole)/(0.05029 L)
= 2.1674 mol/L
