Given the balanced equation
C10H8 + 12O2 --> 10CO2 + 4H2O
In order to find the mass in grams of CO2 that can be produced from 25.0 moles of C10H8, we must convert from moles to mass (mol --> mass conversion).
25.0 mol C10H8 * 10 molecules CO2 * 44.01g CO2 = 1.1025x104 (11025)g CO2
------------------------- 1 molecule C10H8
The combustion of butane is 2 C4H10 + 13 O2 = 8 CO2 + 10 H2O. Every two moles of C4H10 can produce 10 moles of water, so given 12.5 moles of butane, 62.5 moles of water are made.
Two, this is due to the oxygen being the limiting element to form water. If this compound was split to form water would leave C2H4 which is Ethene.
After burning with oxygen: six
If the reaction is complete combustion, all of the hydrogen in C3H8 will be converted to water, with formula H2O. Therefore, 4 moles of water will be produced.
15
26
6 moles COULD be produced
7.30 C2H6O (6 moles H/1 mole C2H6O)(6.022 X 1023/1 mole H) = 2.64 X 1025 atoms of hydrogen =====================
746
1,4 moles of CO are produced.
No moles of oxygen are produced by complete combustion of propane. Oxygen is CONSUMED, not produced. For combustion of 4 moles of propane, it will use 20 moles of oxygen.
Balanced equation. C2H6O + 3O2 --> 2CO2 + 3H2O 0.274 moles C2H6O (2 moles CO2/1 mole C2H6O) = 0.548 moles carbon dioxide produced ============================
6 moles COULD be produced
10,55 moles of water are obtained.
7.30 C2H6O (6 moles H/1 mole C2H6O)(6.022 X 1023/1 mole H) = 2.64 X 1025 atoms of hydrogen =====================
Sulfuric acid is not obtained from water.
6,49 moles of water are obtained.
0 moles
Two moles of water are produced.
746
9
two moles
.003 mols .003 mols