Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Câu 5:
PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)
Ta có: \(n_{Fe}=\dfrac{16,8}{56}=0,3\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{H_2}=0,3\left(mol\right)\\n_{HCl}=0,6\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}V_{H_2}=0,3\cdot22,4=6,72\left(l\right)\\m_{HCl}=0,6\cdot36,5=21,9\left(g\right)\end{matrix}\right.\)
Câu 8 :
a) \(n_{Cu}=\dfrac{6,4}{64}=0,1\left(mol\right)\)
\(n_{Al}=\dfrac{54}{27}=2\left(mol\right)\)
b) \(V_{CO_2}=0,15.22,4=3,36\left(l\right)\)
\(V_{N_2}=0,3.22,4=6,72\left(l\right)\)
c) \(n_{hh}=n_{CO_2}+n_{H_2}=\dfrac{0,22}{44}+\dfrac{0,02}{2}=0,015\left(mol\right)\)
\(V_{hh}=0,015.22,4=0,336\left(l\right)\)
Câu 9
a) \(m_N=0,3.14=4.2\left(g\right)\)
\(m_{Cl}=0,4.35,5=14,2\left(g\right)\)
\(m_O=5.16=80\left(g\right)\)
b) \(m_{N_2}=0,2.28=5,6\left(h\right)\)
\(m_{Cl_2}=0,3.71=21,3\left(g\right)\)
\(m_{O_2}=4.32=128\left(g\right)\)
c) \(m_{Fe}=0,12.56=6,72\left(g\right)\)
\(m_{Cu}=3,15.64=201,6\left(g\right)\)
\(m_{H_2SO_4}=0,85.98=83,3\left(g\right)\)
\(m_{CuSO_4}=0,52.160=83,2\left(g\right)\)
nNaOH = 16/40 = 0,4 (mol)
PTHH: 2NaOH + SO2 -> Na2SO3 + H2O
Mol: 0,4 ---> 0,2 ---> 0,2
VSO2 = 0,2 . 22,4 = 4,48 (l)
mNa2SO3 = 0,2 . 126 = 25,2 (g)
a)
\(m_H=\dfrac{2,04.98}{100}=2\left(g\right)=>n_H=\dfrac{2}{1}=2\left(mol\right)\)
\(m_S=\dfrac{32,65.98}{100}=32\left(g\right)\) => \(n_S=\dfrac{32}{32}=1\left(mol\right)\)
\(m_O=\dfrac{65,31.98}{100}=64\left(g\right)=>n_O=\dfrac{64}{16}=4\left(mol\right)\)
=> CTHH: H2SO4
b)
nH = 2.2 = 4(mol)
nS = 1.2 = 2(mol)
nO = 4.2 = 8 (mol)
Bài 5:
n\(Na_2ZnO_2\) = \(\dfrac{8,58}{143}=0,06mol\)
n\(Mg_3\left(PO_4\right)_2\) =\(\dfrac{7,86}{262}=0,03mol\)
\(n_{CuSO_4}=\dfrac{7,42}{160}=0,046375\approx0,046mol\)
\(n_{CO_2}=\dfrac{26,88}{22,4}=1,2mol\)
\(n_{NO}=\dfrac{20,16}{22,4}=0,9mol\)
\(n_{NO_2}=\dfrac{16,8}{22,4}=0,75mol\)
\(n_{CuSO_4}=\dfrac{1,2.10^{23}}{6.10^{23}}=0,2mol\)
\(n_{Fe}=\dfrac{4,8.10^{23}}{6.10^{23}}=0,8mol\)
n\(Na_3PO_4\) = \(\dfrac{15,3.10^{23}}{6.10^{23}}=2,55mol\)
\(C_nH_{2n+2}+\dfrac{3n-1}{2}O_2\underrightarrow{^{to}}nCO_2+\left(n+1\right)H_2O\)
\(2C_nH_{2n+2}+\left(3n+1\right)O_2\rightarrow2nCO_2+2\left(n+1\right)H_2O\)