Calculate the percentage of water of crystallisation in MgSO⁴ 7H²O

2 answers:

7 0

Answer:

Formula of EPSOM salt = MgSO4.7H2O

molecular mass of MgSO4.7H2O = atomic mass of Mg + atomic mass of S + 4 × atomic mass of O + 7 { 2 × atomic mass of H + atomic mass of O }

= 24 + 32 + 4× 16 + 7{ 2 × 1 + 16 } g/mol

= (24 + 32 + 64+ 126 ) g/mol

= 246 g/mol

molecular mass of total water = 7 × ( 2× atomic mass of H + atomic mass of O )

= 7 × 18 = 126 g/mol

now ,

% mass of H2O in EPSOM salt = {total molar mass of H2O/molar mass of Epsom salts }× 100

= {126/246 } × 100

= 12600/246

= 51.21 %

Explanation:

i have done it hope it helps

6 0

I think that to my calculations it should equal to 51.20 I am not sure but I tried x

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algol13

Im on the same test.

5 0

3 years ago

romanna [79]

Answer:

Coefficient of $H^{+}(aq)$ is more than 4

Explanation:

Oxidation: $Sn^{2+}(aq)\rightarrow Sn^{4+}(aq)$

Reduction: $Cr_{2}O_{7}^{2-}(aq)\rightarrow Cr^{3+}(aq)$

Balance Cr: $Cr_{2}O_{7}^{2-}(aq)\rightarrow 2Cr^{3+}(aq)$
Balance O and H in acidic medium: $Cr_{2}O_{7}^{2-}(aq)+14H^{+}(aq)\rightarrow 2Cr^{3+}(aq)+7H_{2}O(l)$
Balance charge: $Cr_{2}O_{7}^{2-}(aq)+14H^{+}(aq)+6e^{-}\rightarrow 2Cr^{3+}(aq)+7H_{2}O(l)$ .......(2)