Answer:
12.8 g of 
must be withdrawn from tank
Explanation:
Let's assume gas inside tank behaves ideally.
According to ideal gas equation- 
where P is pressure of , V is volume of , n is number of moles of , R is gas constant and T is temperature in kelvin scale.
We can also write, 
Here V, T and R are constants.
So, 
ratio will also be constant before and after removal of from tank
Hence, 
Here, 
and 
So, 
So, moles of must be withdrawn = (0.66 - 0.26) mol = 0.40 mol
Molar mass of = 32 g/mol
So, mass of must be withdrawn = 
Answer:
1. 0.0154mole of PbS
2. Double displacement reaction
Explanation:
First, let write a balanced equation for the reaction. This is illustrated below:
Pb(CH3COO)2 + H2S —> PbS + 2 CH3COOH
Molar Mass of Pb(CH3COO)2 = 207 + 2(12 + 3 + 12 + 16 +16) = 207 + 2(59) = 207 + 118 = 325g
Mass of Pb(CH3COO)2 = 5g
Number of mole = Mass /Molar Mass
Number of mole of Pb(CH3COO)2 = 5/325 = 0.0154mole
From the equation,
1mole of Pb(CH3COO)2 produced 1mole of PbS.
Therefore, 0.0154mole of Pb(CH3COO)2 will also produce 0.0154mole of PbS
2. The name of the reaction is double displacement reaction since the ions in the two reactants interchange to form two different products
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