M Na2CO3 solution (the 25 mL) is
mols/L = 0.04935/0.025 = 1.974 M Na2CO3.
The 1.974M Na2CO3 now reacts with 0.84N H2SO4(0.84N = 0.42M). How much Na2CO3. That's 125 g of the solution. The density of that solution is 1.25 g/mL. volume = mass/density = 125g/1.25 g/mL = 100 mL Na2CO3.
Na2CO3 + H2SO4 ==> Na2SO4 + H2O + CO2
mols Na2CO3 = 0.100L x 1.974M = 0.1974.
mols H2SO4 = the same
M H2SO4 = mols H2SO4/L H2SO4. You have mols (0.1974) and you have M(0.84N or 0.42M). Solve for L H2SO4 That's 0.1974 mols/0.42 M = 0.470 L = 470 mL.
25ml of a solution of na2c03 having a sepcific gravity of 1.25g solution of hcl containing 109.5g of the acid per litre for complete neutralization. Calculate the volume of 0.84N h2s04 that will be compltetly neutralized by 125g of na2c03 solution.
In first part I understand
109.5g/36.5g/mol = 3M
And then 3M x 0.0329L = 0.0987/2 0.04935
In second I do not still understand
How do get 470ml
Step by step explain
2 answers
a 5 g mixture of natural gas containing ch4 and c2h4 was burnt in excess of oxygen yielding 14.5 g of co2 and h2o.What is the weight of ch4 and co2 in the mixture.