Starting material/product |
Molecular weight |
Amount taken |
moles |
Equivalent |
SrCl2 |
158.53 |
5.80 gm |
0.0365 mol |
1.729 |
Li3PO4 |
115.79 |
2.45 gm |
0.0211 mol |
1.00 |
Sr3(PO4)2 |
452.8027 |
Li3PO4 is the limiting reagent in the reaction
115.79 gm of Li3PO4 gives 452.8027 gm of Product
Therefore 2.45 gm of Li3PO4 will give x gm of product
Theoretical yield of the reaction = (2.45*452.8027)/115.79 = 9.58 gm
c. (10 pts.) Using the equation: 3 S-Cl2 (aq) + 2 Li3PO4 (aq) → Sr3(PO4)2 (8)...
3SrCl2(aq) + 2Li3PO4(aq) → Sr3(PO4)2(s) + 6LiCl(aq) What volume of a 1.55 M SrCl2 solution will completely react with 1.50 L of a 1.77 M Li3PO4 solution according to the balanced chemical equation above? You can assume 100% reaction progress.
Question 5 2 pts Consider what would happen if aqueous solutions of strontium chloride and lithium phosphate were combined. Match each of the items below to the correct description of that item in this chemical process. Item 1: 3 Sr2+ (aq) + 2 PO43- (aq) --> Sr3(PO4)2 (s) Item 2: Lit and Cl" Item 3: 3 Sr2+ (aq) + 6 CI" (aq) + 6 LI+ (aq) + 2 PO43- (aq) --> Sr3(PO4)2 (s) +6 Lit (aq) + 6 Cl(aq) Item...
Consider the following unbalanced equation: Ca(ClO3)2(aq) + Li3PO4(aq) → Ca3(PO4)2(s) + LiClO3(aq) If 38.3 moles of Li3PO4(aq) reacts with an excess of Ca(ClO3)2(aq), what is the theoretical yield of LiClO3(aq) in moles?
First determine the actual yield, moles of Sr3(PO4)2 formed. From the actual yield, theoretical moles of limiting reactant in sample: a) Theoretical mass of limiting reactant b) Mass percent of limiting reactant in sample From the actual yield, mass of excess reactant sample, what is the mass percent of excess in sample? Beaker Sample II Unknown mixture NaPO412H2O * SrCl2* 6H2O Mass of mixture and container 113.4904g Mass of container 111.5575g Mass of mixture added to Beaker II 1.9329g Formula...
2. ( pls.) Limiting Reactant and Theoretical Yield (7 pts.) 2Na (s) + Cl2 (g) → 2NaCl (s) You are given 84.9 g Na and 53.2 g Cl2. Find the limiting reactant, and calculate the theoretical yield in grams from the reaction shown above. (Box your answers!) Na = 22.990 g/mol; CI = 35.453 g/mol PACA
Consider the following unbalanced equation: Ca(ClO3)2(aq) + Li3PO4(aq) → Ca3(PO4)2(s) + LiClO3(aq) If 63.2 moles of Ca(ClO3)2(aq) and 22.3 moles of Li3PO4(aq) are allowed to react to produce 40.4 moles of LiClO3(aq), what is the percent yield of the reaction?\ 47.5% 63.8% 73.1% 60.4% 84.0%
Consider the balanced equation for the following reaction: 3Ca(ClO3)2(aq) + 2Li3PO4(aq) → Ca3(PO4)2(s) + 6LiClO3(aq) How much excess reactant remains in the reaction if 44.1 grams of Ca(ClO3)2 reacts with 55.3 grams of Li3PO4?
2. (7 pts.) Limiting Reactant and Theoretical Yield (7 pts.) 2Na (s) + Cl2 (g) - 2NaCl (s) You are given 84.9 g Na and 53.2 g Cl2. Find the limiting reactant, and calculate the theoretical yield in grams from the reaction shown above. (Box your answers!) Na = 22.990 g/mol; CI = 35.453 g/mol 22.990 mo lng & Appoy angl nach 229909x imoina Trimorol mol Nad mol 8Hana 2 molNG =0.265 35.453 g/mol x Imola x 2 mol Nach...
Consider the following balanced equation: 3Ca(ClO3)2(aq) + 2Li3PO4(aq) → Ca3(PO4)2(s) + 6LiClO3(aq) If 41.0 moles of Li3PO4(aq) reacts with an excess of Ca(ClO3)2(aq) to produce 61.9 moles of LiClO3(aq), what is the percent yield of the reaction?
Consider the following balanced equation: 3Ca(ClO3)2(aq) + 2Li3PO4(aq) → Ca3(PO4)2(s) + 6LiClO3(aq) If 15.8 moles of Ca(ClO3)2(aq) reacts with an excess of Li3PO4(aq), and the percent yield is 63.3%, how many moles of LiClO3(aq) will actually be produced