Asked by mandy
Solution A:50cm^3 of 1.0mol dm^-3 hydrochloric acid,HCl
Solution B:10cm^3 of 2.0mol dm^-3 hydrochloric acid,HCl
(a)Calculate the number of moles of hydrochloric acid,HCL in each solution.
(b)Solution A and B are mixed together.
(i)Calculate the number of moles of hydrochloric acid,HCL in the final solution.
(ii)What is the volume of 2.0mol dm^-3 ammonia,NH3 solution needed to neutralise the final acid solution?
Solution B:10cm^3 of 2.0mol dm^-3 hydrochloric acid,HCl
(a)Calculate the number of moles of hydrochloric acid,HCL in each solution.
(b)Solution A and B are mixed together.
(i)Calculate the number of moles of hydrochloric acid,HCL in the final solution.
(ii)What is the volume of 2.0mol dm^-3 ammonia,NH3 solution needed to neutralise the final acid solution?
Answers
Answered by
DrBob222
A. mols soln A = mol/dm3 x dm3 =?
mols soln B = same thing.
Bi. total mols HCl = mols of A + mols B
Bii.
NH3 + HCl ==> NH4Cl
mols HCl you have.
From the coefficients in the balanced equation, you know mols NH3 = mols HCl
Then mols NH3 = mol/dm3 x dm3. You know mols and mol/dm3 NH3, solve for dm3 NH3.
mols soln B = same thing.
Bi. total mols HCl = mols of A + mols B
Bii.
NH3 + HCl ==> NH4Cl
mols HCl you have.
From the coefficients in the balanced equation, you know mols NH3 = mols HCl
Then mols NH3 = mol/dm3 x dm3. You know mols and mol/dm3 NH3, solve for dm3 NH3.
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