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Sagot :
Answer:
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.
Sodium chloride is a limiting reagent.
284 grams of barium phosphate (excess reactant) are left.
Explanation:
Given:
[tex]6NaCl+Ba_3(PO_4)_2\rightarrow 2Na_3PO_4+3BaCl_2[/tex]
1000 grams of sodium chloride and 2000 grams of barium phosphate.
To find:
The limiting reactant and mass of excess reactant left.
Solution:
Mass of sodium chloride = 1000 g
Moles of sodium chloride =[tex]\frac{1000 g}{58.44 g/mol}=17.11mol[/tex]
Mass of barium phosphate = 2000 g
Moles of barium phosphate =[tex]\frac{2000 g}{601.93 g/mol}=3.323 mol[/tex]
[tex]6NaCl+Ba_3(PO_4)_2\rightarrow 2Na_3PO_4+3BaCl_2[/tex]
According to reaction, 6 moles of sodium chloride reacts with 1 mole of barium phosphate, then 17.11 moles of sodium chloride will react with:
[tex]=\frac{1}{6}\times 17.11mol=2.852 \text{ mol of }Ba_3(PO_4)_2[/tex]
2.852 moles of barium phosphate reacted with 17.11 moles of sodium chloride.
But according to the question, we have 3.323 moles of barium phosphate which is more than 2.852 moles.
This indicates that barium phosphate is an excessive reagent and sodium chloride is a limiting reagent.
Moles of barium phosphate left unreacted:
[tex]=3.323 mol - 2.852 mol = 0.471 mol[/tex]
Mass of 0.471 moles of barium phosphate:
[tex]=0.471 mol\times 601.93 g/mol=283.51\approx 284 g[/tex]
Sodium chloride is a limiting reagent.
284 grams of barium phosphate (excess reactant) are left.
Learn more about limiting reagents here:
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