3.5 g Na3PO4 6.40 g Ba(NO3)2 Ba3(PO4)2 gram

3.5 g Na3PO4 6.40 g Ba(NO3)2 Ba3(PO4)2 gram

 

 

If 3.50 g of Na3PO4 and 6.40 g of Ba(NO3)2 are added together,

how many grams of Ba3(PO4)2 will be produced?

Which is the limiting reactant?

 

 

Suppose that a solution containing 3.50 g of sodium phosphate (Na3PO4)

is mixed with a solution containing 6.40 g of barium nitrate (Ba(NO3)2).

Calculate the theoretical yield of barium phosphate (Ba3(PO4)2) during this reaction.

2Na3PO4 + 3Ba(NO3)2 → Ba3(PO4)2 + 6NaNO3

 

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Na3PO4의 몰질량 = 163.94 g/mol

3.5 g / (163.94 g/mol) = 0.02135 mol Na3PO4

( 참고 https://ywpop.tistory.com/7738 )

 

 

 

Ba(NO3)2의 몰질량 = 261.34 g/mol

6.40 / 261.34 = 0.02449 mol Ba(NO3)2

 

 

 

Na3PO4 : Ba(NO3)2 = 2 : 3 계수비(= 몰수비) 이므로,

Na3PO4 : Ba(NO3)2 = 2 : 3 = 0.02135 mol : ? mol

? = 3 × 0.02135 / 2 = 0.03203 mol Ba(NO3)2

---> Ba(NO3)2는 이만큼 없다, 부족하다?

---> Ba(NO3)2가 한계 반응물.

( 참고: 한계 반응물 https://ywpop.tistory.com/3318 )

 

 

 

Ba(NO3)2 : Ba3(PO4)2 = 3 : 1 = 0.02449 mol : ? mol

? = 0.02449 / 3 = 0.008163 mol Ba3(PO4)2

 

 

 

Ba3(PO4)2의 몰질량 = 601.92 g/mol

0.008163 mol × (601.92 g/mol) = 4.913 g Ba3(PO4)2

 

 

 

 

[키워드] Na3PO4 + Ba(NO3)2 → Ba3(PO4)2 + NaNO3, Ba(NO3)2 + Na3PO4, Ba3(PO4)2 침전 기준문서