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natima [27]
3 years ago
15

4 % չայրվող խառնուկներ պարունակող 500 գ ծծմբի այրումից ստացվել է 268,8 լ (ն. պ.) ծծմբի (IV) օքսիդ: Որքա՞ն է ռեակցիայի ելքը (%) .

Chemistry
1 answer:
djverab [1.8K]3 years ago
8 0

see we cannot understand this language

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If 42.8 mL of 0.204 M HCl solution is needed to neutralize a solution of Ca(OH)2, how many grams of Ca(OH)2 must be in the solut
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Hey There!

At neutralisation moles of H⁺ from HCl  = moles of OH⁻ from Ca(OH)2  so :

0.204 * 42.8 / 1000  => 0.0087312 moles

Moles of Ca(OH)2 :

2 HCl + Ca(OH)2 = CaCl2 + 2 H2O

0.0087312 / 2 => 0.0043656 moles (  since each Ca(OH)2 ives 2 OH⁻ ions )

Therefore:

Molar mass Ca(OH)2 = 74.1 g/mol

mass = moles of Ca(OH)2 * molar mass

mass =  0.0043656 * 74.1

mass = 0.32 g of Ca(OH)2


Hope that helps!

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