Answer:
The precipitate is CuS.
Sulfide will precipitate at [S2-]= 3.61*10^-35 M
Explanation:
<u>Step 1: </u>Data given
The solution contains 0.036 M Cu2+ and 0.044 M Fe2+
Ksp (CuS) = 1.3 × 10-36
Ksp (FeS) = 6.3 × 10-18
Step 2: Calculate precipitate
CuS → Cu^2+ + S^2- Ksp= 1.3*10^-36
FeS → Fe^2+ + S^2- Ksp= 6.3*10^-18
Calculate the minimum of amount needed to form precipitates:
Q=Ksp
<u>For copper</u> we have: Ksp=[Cu2+]*[S2-]
Ksp (CuS) = 1.3*10^-36 = 0.036M *[S2-]
[S2-]= 3.61*10^-35 M
<u>For Iron</u> we have: Ksp=[Fe2+]*[S2-]
Ksp(FeS) = 6.3*10^-18 = 0.044M*[S2-]
[S2-]= 1.43*10^-16 M
CuS will form precipitates before FeS., because only 3.61*10^-35 M Sulfur Ions are needed for CuS. For FeS we need 1.43*10^-16 M Sulfur Ions which is much larger.
The precipitate is CuS.
Sulfide will precipitate at [S2-]= 3.61*10^-35 M
Answer:
2.03125g of acetylene
Explanation:
First thing's first, we have to write out the balanced chemical equation;
CaC2(s) + 2H2O(l) → Ca(OH)2(aq) + C2H2(g)
Water is in excess, so CAC2 is our limiting reactant. i.e it determines the amount of product that would be formed.
1 mol of CaC2 produces 1 mol of C2H2
In terms of mass;
Mass = Number of moles * Molar mass
where the molar mass of the elements are;
Ca = 40g/mol
C = 12g/mol
H = 1g/mol
CaC2 = 40+ (2*12) = 64g/mol
C2H2 =( 2 * 12) + ( 2 * 1) = 26g/mol
64g (1 * 64g/mol) of CaC2 produces 26g ( 1mol * 26g/mol) of C2H2
5g would produce x?
64 = 26
5 = x
Upon solving for x we have;
x = (5 * 26) / 64
x = 2.03125g
The balanced equation is 2HgO --> 2Hg + O2
The answer to this question is 159.609 g/mol
