Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) 2Na+2H2O--->2NaOH+H2
n\(_{Na}=\frac{13,8}{23}=0,6\left(mol\right)\)
Theo pthh
n\(_{H2}=\frac{1}{2}n_{Na}=0,3\left(mol\right)\)
V\(_{H2}=0,3.22,4=6,72\left(l\right)\)
b) Theo pthh
n\(_{NaOH}=n_{Na}=0,6\left(mol\right)\)
m\(_{NaOH}=0,6.40=24\left(g\right)\)
c) C\(_{M\left(NaOH\right)}=\frac{0,6}{0,1}=6\left(M\right)\)
\(\text{Na + H2O -> NaOH + 1/2H2}\)
\(\text{a) Ta có: n Na=13,8/23=0,6 mol}\)
Theo ptpu: nH2=1/2 nNa=0,3 mol
\(\Rightarrow\text{ V H2=0,3.22,4=6,72 lít}\)
b) Theo ptpu: nNaOH=nNa=0,6 mol
\(\Rightarrow\text{mNaOH=0,6.40=24 gam}\)
\(\text{c) Ta có V dung dịch sau phản ứng=100 ml =0,1 lít}\)
\(\Rightarrow\text{CM NaOH =nNaOH/V dung dịch=0,6/0,1=6M}\)
a, \(HCl+NaOH\rightarrow NaCl+H_2O\)
b, \(n_{HCl}=0,3.2=0,6\left(mol\right)\)
\(n_{NaOH}=n_{NaCl}=n_{HCl}=0,6\left(mol\right)\)
\(\Rightarrow m_{ddNaOH}=\dfrac{0,6.40}{20\%}=120\left(g\right)\)
c, \(m_{NaCl}=0,6.58,5=35,1\left(g\right)\)
a, \(HCl+NaOH\rightarrow NaCl+H_2O\)
b, \(n_{HCl}=0,3.2=0,6\left(mol\right)\)
Theo PT: \(n_{NaOH}=n_{NaCl}=n_{HCl}=0,6\left(mol\right)\)
\(\Rightarrow m_{NaOH}=0,6.40=24\left(g\right)\)
\(\Rightarrow m_{ddNaOH}=\dfrac{24}{20\%}=120\left(g\right)\)
c, \(m_{NaCl}=0,6.58,5=35,1\left(g\right)\)
Bài 12 :
Oxit axit :
Khí cacbonic : \(CO_2\)
Khí sunfuro : \(SO_2\)
Oxit bazo :
Sắt (III) oxit : \(Fe_2O_3\)
Axit :
Axit clohidric : \(HCl\)
Axit photphoric : \(H_3PO_4\)
Bazo :
Natri hidroxit : \(NaOH\)
Nhôm hidroxit : \(Al\left(OH\right)_3\)
Sắt (III) hidroxit : \(Fe\left(OH\right)_3\)
Muối :
Muối ăn : \(NaCl\)
Kali cacbonat : \(K_2CO_3\)
Canxi sunfat : \(CaSO_4\)
Natri photphat : \(Na_3PO_4\)
Natri hidrosunfua : \(NaHS\)
Canxi hidrocacbonat : \(Ca\left(HCO_3\right)_2\)
Natri đihidrophotphat : \(NaH_2PO_4\)
Magie photphat : \(Mg_3\left(PO_4\right)_2\)
Kẽm nitrat : \(Zn\left(NO_3\right)_2\)
Chúc bạn học tốt
H2SO4 + 2NaOH \(\rightarrow\) Na2SO4 + H2O
a) Ta có: m H2SO4=120.10%=12 gam \(\rightarrow\) nH2SO4=\(\frac{12}{98}\) mol
Theo ptpu: nNa2SO4=nNa2SO4=12/98 mol
\(\rightarrow\)mNa2SO4=\(\frac{12}{98}\) .(23.2+96)=17,39 gam
b)Ta có: nNaOH=2nH2SO4=\(\frac{12}{49}\) mol
\(\rightarrow\) V NaOH=\(\frac{12}{49}\) :2=\(\frac{6}{49}\) lít
a, \(CuCl_2+2NaOH\rightarrow2NaCl+Cu\left(OH\right)_2\)
b, \(n_{NaOH}=\dfrac{10}{40}=0,25\left(mol\right)\)
Theo PT: \(n_{CuCl_2}=n_{Cu\left(OH\right)_2}=\dfrac{1}{2}n_{NaOH}=0,125\left(mol\right)\)
\(\Rightarrow m_{Cu\left(OH\right)_2}=0,125.98=12,25\left(g\right)\)
c, \(C_{M_{CuCl_2}}=\dfrac{0,125}{0,1}=1,25\left(M\right)\)
\(n_{NaOH}=\dfrac{10}{40}=0,25\left(mol\right)\)
PTHH:
\(CuCl_2+2NaOH\rightarrow Cu\left(OH\right)_2+2NaCl\)
0,125 0,25 0,125 0,25
\(m_{Cu\left(OH\right)_2}=0,125.98=12,25\left(g\right)\)
\(C_{M\left(CuCl_2\right)}=\dfrac{0,125}{0,1}=1,25\left(M\right)\)
\(a.Na_2O+H_2O\rightarrow2NaOH\\ b.n_{Na_2O}=\dfrac{31}{62}=0,5mol\\ n_{NaOH}=0,5.2=1mol\\ m_{NaOH}=1.40=40g\)