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camimilligan114
07.02.2021 •
Chemistry
NaCl
+
Ba3(PO4)2
Na3PO4
+
BaCl2
1000 grams of sodium chloride is combined with 2000
grams of barium phosphate.
a)What is the limiting reactant?
b) How many grams of excess reactant are left?
Solved
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Ответ:
a) Sodium Chloride is the limiting reactant.
b)285g of Ba3(PO4)2
Explanation:
a) 1000g Nacl x (1mol of Nacl/58.5g) x (2 Mol Na3PO4/6 mol of NaCl)= 5.70
- 2000g Ba3 (PO4)2 X 1 mol of Ba3 (PO4)2/601.9 x 2 mol of Na3PO4/1 mol of Ba3 (PO4)2= 6.65
Hence; NaCl is the limiting Reactant.
b) 1000g Nacl x (1mol of Nacl/58.5g) x 1 mol of Ba3 (PO4)2/ 6 mol of NaCl x601.9g/1 mol= 1715g Ba3(PO4)2 USED.
2000g- 1715g=285g of Ba3(PO4)2 Left.
Ответ:
So basically you use the ratio they have given you. You either multiply by 2 if image is enlarged or divide by 2 if it is reduced. (Keep in mind I got the 2 from your ratio : 1inch : 2inches) Therefore the find the original width, you multiply 10 by 2, giving you 20 inches.