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$a/$

$MCO_3+H_2SO_4 \to MSO_4+CO_2+H_2O$

$m_{ddH_2SO_4}=\frac{2100}{(M+60).24,5}(g)$

$m_{dd}=\frac{2100.98}{(M+60).24,5}-\frac{21.44}{M+60}(g)$

$C\%=\frac{\frac{21}{M+60}.(M+96)}{\frac{2100.98}{(M+60).24,5}-\frac{21.44}{M+60}+21}=27,27\%$

$\Rightarrow M=24(Mg)$

$b/$

$n_{MgSO_4}=0,25(mol)$

$m_{dd}=\frac{7476}{24+60}+21=110(g)$

$m_{MgSO_4}=110-12,3=97,7(g)$

$n_{MgSO_4(dd)}=\frac{97,7.24,6\%}{120}=0,2(mol)$

$n_{MgSO_4(tách)}=0,25-0,5=0,05(mol)$

$M_{MgSO_4.nH_2O}=\frac{12,3}{0,05}=246(g/mol)$

 $\Rightarrow 120+18n=246$

$\to n=7$

$\to MgSO_4.7H_2O$

$c/$

 Ở $80^{0}C$, $100gH_2O$ hoà tan tối đa $54,8gMgSO_4$ $\to$ $154,8g$dd

                            ?________________________________________________1548

$m_{H_2O}=\frac{1548.100}{154,8}=1000(g)$

$m_{MgSO_4(bd)}=\frac{1458.54,8}{15,8}=548(g)$

 Ở $20^{0}C$, $100gH_2O$ hoà tan tối đa $35,1gMgSO_4$ $\to$ $135,1g$dd

                      1000__________________________________?

$m_{MgSO_4(hoà tan)}=\frac{1000.35,1}{100}=351(g)$

$m_{MgSO_4(tách)}=548-351=197(g)$

MCO3 + H2SO4 -> MSO4 + H2O + CO2

a               a                a                      a

 m dd H2SO4 = 98a x 100/24.5 =  400a

m dd sau pư = 400a + 21 - 44a = 356a + 21

C% MSO4 = (M + 96)a x 100%/(356a+21) = 27.27%

a = 21/(M+60)

=> M = 24 (Mg)

m dd MgSO4 = 110g

m dd sau khi làm lạnh = 97.7g

C bão hòa = 24.56% => n MgSO4 bh = 0.2 mol

=> n MgSO4 tt = 0.25 - 0.2 = 0.05 mol

=> M tt = 12.3 / 0.05 = 246 dvC

=> 120 + 18n = 246 => n =7.

=> MgSO4.7H2O

c)

S 80 = m MgSO4 (1) x 100 / m H2O

=> 54.8 = m MgSO4 (1) x 100 / (1548 - m MgSO4 (1))

=> m MgSO4 (1) = 548g

gọi nMgSO4.7H2O tách ra = b mol

S 20 = m MgSO4 (2) x 100 / m H2O

=> 35.1 = (548 - 120b) x 100 / (1000 - 126b)

-> m MgSO4.7H2O = 126b =...