From the chemical formula, 1 formula unit of KAl (SO4)2.12H2O
encompasses 1 atom of Al = 4 * 2 atoms of O in KAl (SO4)2 + 12 atoms of O in
12H2O which is equal to 20 atoms of O.
So, if you have 1.3 × 10^21 Al atoms, you have 20 * 1.3 × 10^21 O atoms will
now be equal to 2.6 * 10^22 atoms of O.
Answer:
There will be 143,67g CO2 produced
Explanation:
2 C6H6 + 15 O2 → 12 CO2 + 6 H2O
(42,5 g C6H6) / (78.1124 g C6H6/mol) = 0.54408775 mole C6H6
(113.1 g O2) / (31.9989 g O2/mol) = 3.534496 moles O2
0.54408775 mole of C6H6 would react completely with 0.54408775 x (15/2) = 4.080658 mole O2, but there is more O2 present than that, so O2 is in excess and C6H6 is the limiting reactant.
(0.54408775 mol C6H6) x (12/2) x (44.0096 g/mol) = 143.67 g CO2
I hope this helps and if not I’m super sorry that I couldn’t do more!
<h3>GREETINGS!</h3><h2>_____________________________________</h2><h2><u>
Answer</u>
:</h2>
N and C
<h2>_____________________________________</h2><h3><u>WHY OPTION A?</u></h3>
Because Lithium and Sodium belong to group 1A whereas Magnesium is from group 2A, which are S block elements and they form ionic bond.
<h2>_____________________________________</h2><h3 /><h3>HOPE IT HELPS!</h3>
