The moles of butane gas and oxygen gas reacted if 2.50 moles of H2O is produced is calculated as below
the equation for reaction
2C4H10 +13 O2 = 8CO2 +10 H2O
the moles of butane (C4H10) reacted calculation
by use of mole ratio between C4H10: H2O which is 2 : 10 the moles of C4H10= 2.50 x2/10 = 0.5 moles of C4H10 reacted
The moles of O2 reacted calculation
by use of mole ratio between O2 : H2O which is 13:10 the moles of O2
= 2.50 x 13/10= 3.25 moles of O2 reacted
<span>83.9%
First, determine the molar masses of Al(C6H5)3 and C6H6. Start by looking up the atomic weights of the involved elements.
Atomic weight aluminum = 26.981539
Atomic weight carbon = 12.0107
Atomic weight hydrogen = 1.00794
Molar mass Al(C6H5)3 = 26.981539 + 18 * 12.0107 + 15 * 1.00794 = 258.293239 g/mol
Molar mass C6H6 = 6 * 12.0107 + 6 * 1.00794 = 78.11184 g/mol
Now determine how many moles of C6H6 was produced
Moles C6H6 = 0.951 g / 78.11184 g/mol = 0.012174851 mol
Looking at the balanced equation, it indicates that 1 mole of Al(C6H5)3 is required for every 3 moles of C6H6 produced. So given the number of moles of C6H6 you have, determine the number of moles of Al(C6H5)3 that was required.
0.012174851 mol / 3 = 0.004058284 mol
Then multiply by the molar mass to get the number of grams that was originally present.
0.004058284 mol * 258.293239 g/mol = 1.048227218 g
Finally, the weight percent is simply the mass of the reactant divided by the total mass of the sample. So
1.048227218 g / 1.25 g = 0.838581775 = 83.8581775%
And of course, round to 3 significant digits, giving 83.9%</span>
Answer:
the occur near the equator bc its- bc its warm and they get weak bc it slows them down
Explanation:
Answer:
CaO + SO 3 → CaSO. 4
This is an acid-base reaction (neutralization): CaO is a base, SO 3 is an acid.
Answer:
Sodium
Explanation:
Its neighbor to the bottom (sodium) has similar properties to lithium. Group trends are very strong for the Group 1 (alkali) and Group 2 (alkaline earths) metals. Group 1 elements behave very similarly to each other.
