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Report (any type)/Brief report
22/09/2021
The answer is b) 6 g
Start by writing the balanced chemical equation for this
##NaOH_((aq)) + HCl_((aq)) -> NaCl_((aq)) + H_2O_((l))##
Notice that you have a ##1:1## between sodium hydroxide and hydrochloic acid; this means that a complete would require equal numbers of moles of sodium hydroxide and hydrochloric acid.
Now, you know that the hydrochloric acid solution has a normality of 0.1 N. Normality is simply a measure of reactivity, meaning that it is calculated by taking into account how a substance behaves in a particular reaction.
Hydrochloric acid dissociates in aqueous solution to rpoduce
##HCl_((aq)) -> H_((aq))^(+) + Cl_((aq))^(-)##
The net ionic equation for your reaction will be
##OH_((aq))^(-) + H_((aq))^(+) -> H_2O_((l))##
In your case, a 0.1 N solution means that the hydrochloric acid solution provides 0.1 moles of protons, ##H^(+)##, per liter to the reaction.
Since 1 mole of ##”HCl”## produces 1 mole of ##H^(+)##, the of the solution will be equal to ##”0.1 mol/L”##.
The volume of the solution will be
##1500color(red)(cancel(color(black)(“cm”^3))) * (“1 dm”””^3)/(1000color(red)(cancel(color(black)(“cm”^3)))) = “1.5 dm”””^3##
This means that you can now calculate how many moles of hydrochloric acid took part in the reaction (remember that ##”1 dm”^3 = “1 L”##)
##C = n/V implies n = C * V##
##n_”HCl” = 0.1 “moles”/color(red)(cancel(color(black)(“L”))) * 1.5color(red)(cancel(color(black)(“L”))) = “0.15 moles HCl”##
The aforementioned mole ratio tells you that the number of moles of sodium hydroxide needed to neutralize this many moles of hydrochloric acid is
##n_(NaOH) = n_(HCl) = “0.15 moles”##
To get the mass of sodium hydroxide needed, use its molar mass
##0.15color(red)(cancel(color(black)(“moles”))) * “40.0 g”/(1color(red)(cancel(color(black)(“mole”)))) = color(green)(“6 g”)##
I want to thank your team (Jeff) for finding a new writer when you were unable to reach my existing writer to revise my paper. I appreciate your professionalism and efforts. Thanks
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