Hope this helps you.
The student's lab manual says to mix 50 mL of his 2.0 M CaCl2 solution with 50 mL of a 3.0 M aqueous solution of sodium carbonat
1 answer:
You might be interested in
Answer: D. 19.9 g hydrogen remains.
Explanation:
To calculate the moles, we use the equation:
a) moles of
b) moles of
According to stoichiometry :
1 mole of require 1 mole of
Thus 0.0787 moles of require=
of
Thus is the limiting reagent as it limits the formation of product and
acts as the excess reagent. (10.0-0.0787)= 9.92 moles of
are left unreacted.
Mass of
Thus 19.9 g of remains unreacted.