a: \(CO_2+2NaOH\rightarrow Na_2CO_3+H_2O\)

b: \(n_{CO_2}=\dfrac{3.36}{22.4}=0.15\left(mol\right)\)

\(\Leftrightarrow n_{H_2O}=2\cdot0.15=0.3\left(mol\right)\)

\(\Leftrightarrow n_{NaOH}=0.15\left(mol\right)\)

\(m_{NaOH}=0.15\cdot40=6\left(g\right)\)

a) 2NaOH + CO₂ --> Na₂CO₃ + H₂O

b) \(n_{CO_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

PTHH: 2NaOH + CO₂ --> Na₂CO₃ + H₂O

______0,3<---0,15------->0,15------>0,15

=> m_NaOH = 0,3.40 = 12 (g)

c) m_sp = 0,15.106 + 0,15.18 = 18,6(g)

Bài này anh có làm rồi em nha

\(n_{CO_2}=\dfrac{1,12}{22,4}=0,05mol\\ a)CO_2+2NaOH\rightarrow Na_2CO_3+H_2O\)

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

\(b)C_{M_{NaOH}}=\dfrac{0,1}{0,1}=1M\\ c)m_{Na_2CO_3}=0,05.106=5,3g\)

\(n_{H_2}=\dfrac{0,672}{22,4}=0,03\left(mol\right)\)

a. \(2Al+2NaOH+2H_2O\rightarrow2NaAlO_2+3H_2\)

0,02       0,02        0,02         0,02          0,03

Fe không pứ với dd NaOH

b. \(\%_{m_{Al}}=\dfrac{0,02.27.100}{0,78}=69,23\%\)

=> \(\%_{m_{Fe}}=100-69,23=30,77\%\)

\(n_{NaOH}=0.2\cdot0.5=0.1\left(mol\right)\)

\(n_{CuSO_4}=0.1\cdot2=0.2\left(mol\right)\)

\(2NaOH+CuSO_4\rightarrow Na_2SO_4+Cu\left(OH\right)_2\)

\(0.1.............0.05...............0.05...........0.05\)

\(m_{Cu\left(OH\right)_2}=0.05\cdot98=4.9\left(g\right)\)

\(C_{M_{Na_2SO_4}}=\dfrac{0.05}{0.2+0.1}=0.167\left(M\right)\)

\(C_{M_{CuSO_4\left(dư\right)}}=\dfrac{0.2-0.05}{0.1}=1.5\left(M\right)\)

\(n_{MgCl_2}=0.1\times0,2=0,02\left(mol\right)\)

\(n_{NaOH}=\frac{30}{40}=0,75\left(mol\right)\)

a/ PTHH : MgCl₂ + 2NaOH -----> Mg(OH)₂ + 2NaCl

Ta lập tỉ lệ mol : \(\frac{n_{MgCl_2}\left(\text{đề cho}\right)}{n_{MgCl_2}\left(pt\right)}=\frac{0,02}{1}\)

\(\frac{n_{NaOH}\left(\text{đề cho}\right)}{n_{NaOH}\left(pt\right)}=\frac{0,75}{2}=0,375\)

Suy ra NaOH dư , tính số mol Mg(OH)2 theo số mol của MgCl2

Theo pt thì nMg(OH)2 = nMgCl2 = 0,02 (mol)

=> mMgCl2 = 0,02 x 95 = 1,9 (g)

a, \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

\(H_2+CuO\underrightarrow{t^o}Cu+H_2O\)

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

Theo PT: \(n_{ZnCl_2}=n_{H_2}=n_{Zn}=0,2\left(mol\right)\Rightarrow m_{ZnCl_2}=0,2.136=27,2\left(g\right)\)

c, \(n_{CuO}=\dfrac{20}{80}=0,25\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{0,2}{1}< \dfrac{0,25}{1}\), ta được CuO dư.

Theo PT: \(n_{CuO\left(pư\right)}=n_{H_2}=0,2\left(mol\right)\Rightarrow n_{CuO\left(dư\right)}=0,25-0,2=0,05\left(mol\right)\)

\(\Rightarrow m_{CuO\left(dư\right)}=0,05.80=4\left(g\right)\)

a) nNaOH= 0,15(mol) ; nH3PO4= 0,1(mol)

Ta có: 1< 0,15/0,1 <2

=> Sản phẩm thu được là hỗn hợp 2 muối: NaH2PO4 và Na2HPO4

a) PTHH: NaOH + H3PO4 -> NaH2PO4 + H2O

x____________x__________x(mol)

2 NaOH + H3PO4 -> Na2HPO4 + 2 H2O

y_______0,5y_________0,5y(mol)

b) Ta có hpt:

\(\left\{{}\begin{matrix}x+y=0,15\\x+0,5y=0,1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,05\\y=0,1\end{matrix}\right.\)

=> m(muối)=mNaH2PO4 + mNa2HPO4= 120.0,05+142.0,1.0,5=13,1(g)

c) Vddmuoi= VddNaOH+ VddH3PO4= 0,1+0,1=0,2(l)

=> CMddNaH2PO4= 0,05/0,2=0,25(M)

CMddNa2HPO4=(0,1.0,5)/0,2=0,25(M)