Reaction for precipitation is
Pb(NO3)2(aq) + 2Nacl(aq)→PbCL2(s)+2NaNO3(aq)
The given reaction is
The mol of Nacl required is =2×mol of Pb(NO3)2 which is present.
∴M(Nacl)×V(Nacl) =2×M(Pb(NO3)2 ×V(Pb(NO3)2
o.100M×V(Nacl)=2×0.200 MX75.0ML
V(Nacl = 300mL.
Answer:
I think it's more than 100,000 mold
Answer:
O lowering the temperature of the system
1) Find the number of mols of HCl in 5.2 liters of 4.0M solution:
n = M*V(L) = 4.0 mol/L * 5.2 L = 20.8 mol
2) Find the number of mols of Mg that will react with 20.8 mol of HCl, using the coefficients of the balanced equation
[1mol Mg / 2 mol HCl] * 20.8 mol HCl = 10.4 mol Mg
3) Transform mol to mass using the atomic mass:
10.4 mol Mg * 24.3 g/mol = 252.7 g of Mg.
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