A student was given the task of titrating a 20.mL sample of 0.10MHCl(aq) with 0.10MNaOH(aq). The HCl(aq) was placed in an Erlenmeyer flask. An equation for the reaction that occurs during the titration is given above. (a) According to the equation for the reaction, if the amount of the reactants is halved, how does this affect the amount of H2O(l) produced in the reaction

The amount H2O produced is also halved

Explanation:

A strong acid as HCl reacts with a strong base as NaOH producing water and a salt, NaCl. The reaction is:

HCl + NaOH → NaCl + H₂O

Where 1 mole of HCl reacts per mole of NaOH to produce 1 mole of water.

That means, if the amount of HCl and NaOH is halved, the amount H2O produced is also halved because 1 mole of each reactant produce 1 mole of water

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2