Answer:
The answer to your question is letter c. Ca(NO₃)₂
Explanation:
Data
Ca(NO₃)₂ = 3.4 moles
Li₃PO₄ = 2.4 moles
Reaction
3 Ca(NO₃)₂ + 2 Li₃PO₄ ⇒ 6 LiNO₃ + Ca₃(PO₄)₂
Process
Calculate the proportion theoretical and experimental of reactants and compare these proportions.
Theoretical proportion
Ca(NO₃)₂ / Li₃PO₄ = = 1.5
Experimental proportion
Ca(NO₃)₂ / Li₃PO₄ =
As the experimental proportion is lower than the theoretical proportion we conclude that the amount of Li₃PO₄ increased in the experiment so the limiting reactant is Ca(NO₃)₂.
Answer:
245 mL
Explanation:
Please see the step-by-step solution in the picture attached below.
Hope this answer can help you. Have a nice day!
Answer:
beryllium iodide has a molar mass of 262.821 g mol−1 , which means that 1 mole of beryllium iodide has a mass of 262.821 g . To find the mass of 0.02 moles of beryllium iodide, simply multiply the number of moles by the molar mass in conversion factor form.
Explanation:
So so confused uh jus listen to the other answer aha
Answer:
D
Explanation:
Because it has equal number of elements on each side of the arrow