Answer:
D. CI706
Explanation:
I hope it's it if not I'm stupid
C3H8+3O2--->3CO2+8H
Therefore for every 1:3 there are 3 Carbon dioxides that form. That means find the limiting reactant from the two reactants.
5.5g(1mole C3H8/44.03g of C3H8)=0.1249 moled of C3H8 and if for every one C3H8 we can form three CO2. We can assume 0.3747 miles of CO2 will be produced.
15g of O2(1 mole O2/32g of O2)=0.4685moles O2 and if for every three O2 we can produce three CO2 we may assume a 1:1 ratio.
This means C3H8 will be your limiting reactant. Therefore 0.3747 moles of CO2 will be produced.
0.3747 moles of CO2(48.01 g of CO2/1 mole of CO2)= 17.99 grams of CO2
The answer is the last option
Lustrous
High melting point
Malleable
Ductility
1) Molar mass C8H9NO2
Element Atomic mass # of atoms mass
g/mol g
C 12 8 12*8 = 96
H 1 9 1*9 = 9
N 14 1 14*1 = 14
O 16 2 16*2 = 32
molar mass = 96 + 9 + 14 + 32 = 151 g/mol
2) Number of mols in a tablet
# of moles = mass / molar mass = 0.500 g / 151 g/mol = 0.003311 moles
3) 3 doses * 2 tablets * 0.003311 moles / tablet = 0.020 moles
