PLEASE SHOW WORK When calcium carbonate is added to hydrochloric acid, calcium c

Question

PLEASE SHOW WORK

When calcium carbonate is added to hydrochloric acid, calcium chloride, carbon dioxide, and water are produced.

How many grams of calcium chloride will be produced when 28.0 g of calcium carbonate are combined with 15.0 g of hydrochloric acid?

Which reactant is in excess and how many grams of this reactant will remain after the reaction is complete?

2 PLEASE SHOW WORK

Cryolite, Na3AlF6(s), an ore used in the production of aluminum, can be synthesized using aluminum oxide.

Balance the equation.

If 12.0 kilograms of Al2O3(s), 60.4 kilograms of NaOH(l), and 60.4 kilograms of HF(g) react completely, how many kilograms of cryolite will be produced?

PLEASE SHOW WORK When calcium carbonate is added to hydrochloric acid, calcium chloride, carbon dioxide, and water are produced. How many grams of calcium chloride will be produced when 28.0 g of calcium carbonate are combined with 15.0 g of hydrochloric acid? What is the total mass of the excess reactants left over after the reaction is complete? Which reactants will be in excess? If 12.0 kilograms of Al2O3(s), 60.4 kilograms of NaOH(l), and 60.4 kilograms of HF(g) react completely, how many kilograms of cryolite will be produced? 2 PLEASE SHOW WORK Cryolite, Na3AlF6(s), an ore used in the production of aluminum, can be synthesized using aluminum oxide. Balance the equation Which reactant is in excess and how many grams of this reactant will remain after the reaction is complete?

Explanation / Answer

1) For each mole of calcium carbonate we need 2 of hydrochloric acid.

So we got that 1 mol of CaCO3 = 40,08g+12,00g+(15,99g x3) = 100,05g

1mol of HCl = 1,00g=35,44g= 36,44g ==> 2 mol of HCl = 72,88g

1mol of CaCl2 =40,08g + (35,44g x2) ==> 110,96g

100,05g CaCO3----------72,88g HCl

28,00g CaCO3---------X: 20,39 g HCl

if we have 28,00 g of calcium carbonate we will need 20,93g of hydrochloric acid, but we only have 15,00g so calcium carbonate is in excess.

72,88g HCl -------------------100,05g CaCO3

15,00g HCl-------------------X: 20,59g CaCO3

15g of HCl reacts with 20,59g of CaCO3 so we got 28,00g-20,59g = 7,59g of CaCO3 in excess.

If 72,88g HCl --------(produces)----------110,96g CaCl2

20,39g HCl----------(will produce)---------x:31,04g Cacl2

2) Al2O3(s) + 6NaOH(l) + 12HF(g)  ---> 2Na3AlF6(s) + 9H2O(g)

12kg Al2O3 =1200g

60,4kg NaOH =60400g

60,4kg 12HF= 60400g

1mol of Al2O3 =(26,98g x2) + (15,99g x3) = 101,93g

1mol of HF = 1,00g +18,99g =19,99g ===> 12 mol of HF = 239,88g

1mol of NaOH= 22,98g +15,99g +1,00g = 39,97g ===> 6mol of NaOH = 239,82g

1 mol of Na3AlF6 =(22,98g x3)+26,98g + (18,99g x6) = 209,86g ===> 2 mol of Na3AlF6 = 419,72g

101,93g Al2O3---reacts with------239,82g NaOH

if we have 1200g Al2O3 ------we will need ----x: 2823,34g NaOH

101,93g Al2O3---reacts with------239,88g HF

if we have 1200g Al2O3---we will need----x: 2824,05g HF

So NaOH and HFwill be in excess. 60400g-2823,34g NaOH=57576,66g ==> 57,57kg NaOH (excess)

60400g-2824,05g HF= 57575,95g ==> 57,57kg HF (excess)

Total mass of excess reactants: 57,57kg+57,57kg= 115,14kg

if 101,93g Al2O3---produces-----419,72g of Na3AlF6

1200g Al2O3---will produce-----x: 4941,27g of Na3AlF6

4941,27g of Na3AlF6 = 4,94kg of cryolite.

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.