Create the Equation: How many grams of Aluminum Chloride would be made from 42.7 L of Chlorine gas at STP reacting with 50.0 g o
1 answer:
You might be interested in
Answer:
14.533 grams of solid precipitate of mercury(II) dichromate will form.
Explanation:
Moles of mercury(II) acetate =
Moles of sodium dichromate =
According to reaction , 1 mole of sodium dichromate reacts with 1 mole of mercury(II) acetate , then 0.045906 moles of sodium dichromate will recat with :
of mercury(II) acetate
This means that mercury(II) acetate is present in an excess amount and sodium dichromate is present in limiting amount.So, amount of precipitate will depend upon moles of sodium dichromate.
According to reaction , 1 mole of sodium dichromate gives 1 mole of mercury(II) dichromate , then 0.045906 moles of sodium dichromate will give :
of mercury(II) dichromate
Mass of 0.045906 moles of mercury(II) dichromate:
0.045906 mol × 316.59 g/mol = 14.533 g
14.533 grams of solid precipitate of mercury(II) dichromate will form.
Hello!
answer : c
eg = octahedral, mg = square planar, sp3d2
Hope that helps!
