How many mL of 0.500 M NaOH would you have to add to neutralize the remaining excess HCl(aq) in the reaction?

[匿名]

2017-10-20 10:14 am

You react Mg with HCl as seen below:

Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)

You are given 1.80 grams of Mg(s) ribbon and told house you to use 200.0 mL of 0.900 M HCl(aq), use this information to determine the following:
How many mL of 0.500 M NaOH would you have to add to neutralize the remaining excess HCl(aq) in the reaction?

回答 (2)

wanszeto

2017-10-20 10:37 am

Molar mass of Mg = 24.3 g/mol
Initial no. of moles of Mg = (1.80 g) / (24.3 g/mol) = 0.07407 mol
Initial no. of moles of HCl = (0.900 mol/L) × (200.0/1000 L) = 0.18 mol

Consider the reaction between Mg and HCl :
Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g)
Mole ratio Mg : HCl = 1 : 2
When 0.0741 mol Mg completely reacts, HCl reacted = (0.07407 mol) × 2 = 0.14814 mol
No. of moles of excess HCl = 0.18 - 0.14814 = 0.03186 mol

Consider the reaction between NaOH and excess HCl :
NaOH + HCl → NaCl + H₂O
No. of moles of NaOH needed = No. of moles of excess HCl = 0.0318 mol
Volume of NaOH needed = (0.03186 mol) / (0.500 mol/L) = 0.0637 L = 63.7 mL

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[匿名]

2017-10-20 10:16 am

jrs

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