Answer:
Explanation:
Molecular weight of barium sulphide = 169
500 mL of .192 M barium sulfide = .5 x .192 moles of barium sulphide
= .096 moles of barium sulfide
= .096 x 169 gram of barium sulfide
= 16.22 grams of barium sulfide .
We shall have to add 16.22 gram .
Roots of trees breaking sidewalks, hope this helps
Answer:
2.843 M
Explanation:
Molarity = moles / volume
<u>Milliliters to liters:</u>
500 mL <u>= .500 L</u>
<u>Grams to moles:</u>
MgO molar mass = 16.00 + 24.31 = 40.31 g/mol
57.3 g x 1 mol / 40.31 g = <u>1.421 mol</u>
<u>Molarity:</u>
1.421 mol / .500 L = 2.843 M
Answer: 313.6
Explanation:
To calculate the moles :
According to stoichiometry :
1 mole of require 3 moles of
Thus 2.8 moles of will require= of
Thus is the limiting reagent as it limits the formation of product and is the excess reagent.
As 1 mole of give = 2 moles of
Thus 2.8 moles of give = of
Mass of
Theoretical yield of liquid iron is 313.6 g