lead tin 3.50 g HNO3 1.57 g PbSO4(s)

lead tin 3.50 g HNO3 1.57 g PbSO4(s)

 

 

A 3.50 g ofan alloy which contains only lead and tin is dissolved in hot HNO3.

Excess sulfuric acid is added to this solution and 1.57 g of PbSO4(s) is obtained.

a) Write the net ionic equation for the formation of PbSO4(s).

b) Assuming all the lead in the alloy reacted to form PbSO4,

what was the amount, in grams, of lead and tin in the alloy respectively?

 

 

 

net ionic equation

Pb^2+(aq) + SO4^2-(aq) → PbSO4(s)

 

 

 

PbSO4의 몰질량 = 303.26 g/mol 이므로,

1.57 g / (303.26 g/mol) = 0.005177 mol PbSO4

( 참고 https://ywpop.tistory.com/7738 )

 

= 0.005177 mol Pb

 

 

 

Pb의 몰질량 = 207.20 g/mol 이므로,

0.005177 mol × (207.20 g/mol) = 1.07 g Pb

 

 

 

3.50 – 1.07 = 2.43 g Sn

 

 

 

 

[키워드] lead and tin alloy 기준, Pb Sn 합금 기준