Answer:
12.44 g
Explanation:
2C4H10 + 13O2 = 8CO2 + 10H2O
n(C4H10) = m(C4H10)/M(C4H10) = 4.1 / 58g/mol = 0.0707 mol (excess).
n(O2) = m(O2)/M(O2) = 25.9 / 32g/mol = 0.809 mol (deficiency).
Since the ratio of O2 to octane is 13 : 2 we can divide 0.0707 by 2 to get 0.03535 and divide 0.809 by 13 to get 0.062.
mass of CO2 produced =
M = [0.0707 moles C4H10 x 8 moles CO2] / 2 moles C4H10 x 44 g CO2/mol
M = 0.5656/2 * 44
M = 0.2828 * 44
M = 12.44 of CO2
Oxygen and carbon dioxide
Answer:
Explanation:
Percent composition by element
Element Symbol # of Atoms
Chlorine Cl 2
Calcium Ca 1
Oxygen O 6
Answer: 1.8124 g of 3-nitrophthalhydrazide were recovered.
Explanation:
The balanced chemical reaction will be :
moles of 3-nitrophthalic acid = 
As 1 mole of 3-nitrophthalic acid gives = 1 mole of 3-nitrophthalhydrazide
0.0110 moles of 3-nitrophthalic acid gives = 
mole of 3-nitrophthalhydrazide
mass of 3-nitrophthalhydrazide = 
As the percentage yield is 93% , the mass of 3-nitrophthalhydrazide recovered = 
Therefore 1.8124 g of 3-nitrophthalhydrazide were recovered.
