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junior college 1 | H2 Maths
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Sonia
Sonia

junior college 1 chevron_right H2 Maths chevron_right Singapore

#H2Chem

Good evening! Can someone please help me with this question? I have trouble finding the mass of NaCl, i could only find the moles of Hcl & NaOH and i’m stuck afterwards. Thank you so much

Date Posted: 4 years ago
Views: 280
Eric Nicholas K
Eric Nicholas K
4 years ago
Good evening Sonia! I will have a look later if I have the time.
Eric Nicholas K
Eric Nicholas K
4 years ago
Since 22.4 cm3 of 0.5 mol/dm3 HCl is used,

Number of moles of HCl used
= 0.0224 x 0.5
= 0.0112 mol

Since NaOH reacts with HCl in a 1 : 1 ratio, number of moles of NaOH reacted
= 0.0112 mol

This means that the residual quantity of NaOH before reaction with HCl must have been 0.0112 x 10 = 0.112 mol in the 250 cm3 solution, since there is 0.0112 mol in every 25 cm3.

Initially 100 cm3 of 2.00 mol/dm3 is used.

Number of moles of NaOH at first
= 0.1 x 2.00
= 0.2 mol

This means that 0.2 - 0.112 = 0.088 mol of NaOH must have been reacted with the ammonium chloride.

Equivalent mass of NH4Cl
= 0.088 * (14 + 4 * 1 + 35.5)
= 4.708 g

and as such, mass NaCl in the contaminated 5 g sample
= 5 g - 4.708 g
= 0.292 g
Eric Nicholas K
Eric Nicholas K
4 years ago
Sonia, I will write out a full explanation of my thoughts on this problem later by 2 am after I return home.

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Eric Nicholas K
Eric Nicholas K's answer
5997 answers (Tutor Details)
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Good evening Sonia! Here are my thought processes for this question. I will send in my remaining workings soon. Let me know if you need more explanation for my workings.
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Eric Nicholas K
Eric Nicholas K's answer
5997 answers (Tutor Details)
Good evening Sonia! This second page completes my working for this question. Let me know if my explanations and thought processes are not clear and I will do my best to explain them again.