Answer:
A. m C5H12 = 108.23 g
B. m F2 = 547.142 g
C. m Ca(CN)2 = 71.85 g
Explanation:
- mass (m) = mol (n) × molecular weigth (Mw)
∴ Mw C5H12 = ((12.011)(5)) + ((1.008)(12)) = 72.151 g/mol C5H12
∴ Mw F2 = (18.998)(2) = 37.996 g/mol F2
∴ Mw = Ca(CN)2 = 40.078+((12.011+14.007)(2)) = 92.114 g/mol Ca(CN)2
A. m C5H12 = ( 1.50 mol)×(72.151 g/mol) = 108.23 g C5H12
B. m F2 = (14.4 mol)×(37.996 g/mol) = 547.142 g F2
C. m Ca(CN)2 = (0.780 mol)×(92.114 g/mol) = 71.85 g Ca(CN)2
The postulate of Dalton's atomic theory which is a result of the law of conservation of mass is: Atoms are indivisible particles, which can neither be created nor destroyed in a chemical reaction.
Answer:
0.912 mL
Explanation:
3 Na2S(aq) + 2 FeCl3(aq) → Fe2S3(s) + 6 NaCl(aq)
FeCl3 is the limiting reactant.
Number of moles of iron III sulphide produced= 3.75g/87.92 g/mol = 0.043 moles
Hence actual yield of Iron III sulphide = 0.043 moles
Theoretical yield of Iron III sulphide = actual yield ×100%/ %yield
Theoretical yield of iron III sulphide= 0.043 ×100/75 = 0.057 moles of Iron III sulphide
From the reaction equation,
2moles of iron III chloride produced 1 mole of iron III sulphide
x moles of iron III chloride, will produce 0.057 of iron III sulphide
x= 2× 0.057= 0.114 moles of iron III chloride
But
Volume= number of moles/ concentration
Volume= 0.114/0.125
Volume= 0.912 mL
