1. \(SO_3+H_2O\rightarrow H_2SO_4\)

Gọi a là số mol SO3 cần thêm

\(m_{Dd_{Spu}}=80+80a\left(g\right)\)

Ta có

\(\frac{98a}{80+80a}=24,5\%\)

\(\Rightarrow a=0,25\left(mol\right)\)

\(m_{SO3}=0,25.80=20\left(g\right)\)

2. \(CaO+H_2O\rightarrow Ca\left(OH\right)_2\)

Gọi a là số mol CaO tham gia

\(m_{dd_{Spu}}=150+56a\left(g\right)\)

Ta có

\(\frac{74a}{150+56a}=25\%\)

\(\Rightarrow a=0,625\)

\(m_{CaO}=0,625.56=35\left(g\right)\)

Gọi 

\(n_{SO_3}=a\left(mol\right)\\ \rightarrow m_{dd\left(sau\right)}=600+80a\left(g\right)\\ PTHH:SO_3+H_2O\rightarrow H_2SO_4\\ Mol:a\rightarrow a\rightarrow a\\ m_{H_2SO_4\left(bđ\right)}=24,5\%.600=147\left(g\right)\\ \rightarrow C\%_{H_2SO_4\left(sau\right)}=\dfrac{146+98a}{600+80a}=49\%\\ \Leftrightarrow n_{SO_3}=2,5\left(mol\right)\\ \rightarrow m_{SO_3}=2,5.80=200\left(g\right)\)

Gọi \(n_{SO_3}=a\left(mol\right)\)

\(\rightarrow m_{dd\left(sau.khi.hoà.tan.thêm\right)}=600+80a\left(g\right)\)

\(PTHH:SO_3+H_2O\rightarrow H_2SO_4\)

                a                        a

\(m_{H_2SO_4\left(bđ\right)}=24,5\%.600=147\left(g\right)\\ \rightarrow C\%_{H_2SO_4\left(sau.khi.pha\right)}=\dfrac{147+98a}{600+a}=49\%\\ \Leftrightarrow a\approx1,51\left(mol\right)\\ \rightarrow m_{SO_3}=1,51.80=120,8\left(g\right)\)

Độ tan của NaCl ở 25 độ C là 36 gam, nghĩa là:

100 gam \(H_2O\) hòa tan được 36 gam NaCl:

Suy ra với 80 gam \(H_2O\) sẽ hòa tan được: \(\dfrac{36}{100}.80=28,8\left(g\right)\)

Ta có: \(24\left(g\right)< 28,8\left(g\right)\Rightarrow\) dung dịch NaCl chưa bão hòa

Cần thêm khối lượng NaCl để dung dịch được bảo hòa:

\(28,8-24=4,8\left(g\right)\)

\(n_{H_2}=\dfrac{22,4}{22,4}=1\left(mol\right);n_{Cl_2}=\dfrac{33,6}{22,4}=1,5\left(mol\right)\\ PTHH:H_2+Cl_2\rightarrow\left(as\right)2HCl\\ Vì:\dfrac{1}{1}>\dfrac{1,5}{1}\Rightarrow Cl_2dư\\ n_{HCl\left(thu.được\right)}=75\%.n_{H_2}=75\%.1=0,75\left(mol\right)\\ Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\\ n_{Fe_2O_3}=\dfrac{n_{HCl}}{6}=\dfrac{0,75}{6}=0,125\left(mol\right)\\ \Rightarrow m_{Fe_2O_3}=160.0,125=20\left(g\right)\)

Ai cứu mình với ạ mình đang cần gấp🥲

Na2O+H2O->2NaOH

0,05-------------------0,1

n NaOH=0,1 mol

=>m Na2O=0,05.62=3,1g

$n_{KOH} = \dfrac{11,2}{56} = 0,2(mol)$
$C_{M_{KOH}} = \dfrac{0,2}{0,15} = 1,33M$

\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)

\(PTHH:Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)

(mol)_____0,2____0,2______0,2____0,2__

\(a.V_{H_2}=22,4.0,2=4,48\left(l\right)\)

\(b.m_{ddH_2SO_4}=\dfrac{0,2.98.100}{24,5}=80\left(g\right)\)

\(c.m_{ddspu}=13+80-0,2.2=92,6\left(g\right)\\ \Rightarrow C\%_{ddspu}=\dfrac{0,2.136}{92,6}.100=29,4\left(\%\right)\)

a, \(n_K=\dfrac{3,9}{39}=0,1\left(mol\right)\)

PTHH: 2K + 2H₂O ---> 2KOH + H₂

            0,1---------------->0,1----->0,05

\(m_{ct}=m_{KOH}=0,1.56=5,6\left(g\right)\\ m_{dd}=m_K+m_{H_2O}-m_{H_2}=96,2+3,9-0,05.2=100\left(g\right)\)

\(C\%_{KOH}=\dfrac{5,6}{100}.100\%=5,6\%\\ b,m_{dd}=100+50=150\left(g\right)\\ C\%_{KOH}=\dfrac{5,6}{150}.100\%=3,37\%\)

c, Gọi \(m_{H_2O}=a\left(g\right)\)

\(\Rightarrow C\%_{KOH}=\dfrac{5,6}{100+a}.100\%=2,8\%\\ \Leftrightarrow a=100\left(g\right)\)

d, Gọi \(m_{KOH}=a\left(g\right)\)

\(\Rightarrow C\%_{KOH}=\dfrac{5,6+a}{100+a}.100\%=22,4\%\\ \Leftrightarrow a=21,65\left(g\right)\)