Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a)
4P + 5O2 --to--> 2P2O5
P2O5 + 3H2O --> 2H3PO4
b)
Giả sử trong dd X có chứa \(\left\{{}\begin{matrix}H_3PO_4:x\left(mol\right)\\H_2O:y\left(mol\right)\end{matrix}\right.\)
\(\left\{{}\begin{matrix}n_H=3x+2y\left(mol\right)\\n_O=4x+y\left(mol\right)\end{matrix}\right.\)
=> \(\dfrac{4x+y}{3x+2y}=\dfrac{4}{7}\) => 28x + 7y = 12x + 8y
=> y = 16x
Có: \(C\%=\dfrac{98x}{98x+18y}.100\%=\dfrac{98x}{98x+18.16x}.100\%=25,389\%\)
Câu 5. a) \(SO_2+\dfrac{1}{2}O_2-^{t^o,V_2O_5}\rightarrow SO_3\)
\(n_{SO_3}=n_{SO_2}=0,1\left(mol\right)\)
\(SO_3+H_2O\rightarrow H_2SO_4\)
\(n_{H_2SO_4}=n_{SO_3}=0,1\left(mol\right)\)
=> \(CM_{H_2SO_4}=\dfrac{0,1}{0,2}=0,5M\)
b) \(n_{Zn}=0,05\left(mol\right)\)
\(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
Lập tỉ lệ : \(\dfrac{0,05}{1}< \dfrac{0,1}{1}\)=> Sau phản ứng H2SO4 dư
=> \(m_{H_2SO_4\left(dư\right)}=\left(0,1-0,05\right).98=4,9\left(g\right)\)
Câu 5 . \(n_{Al_2O_3}=0,2\left(mol\right);n_{H_2SO_4}=0,8\left(mol\right)\)
\(Al_2O_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O\)
Lập tỉ lệ: \(\dfrac{0,2}{1}< \dfrac{0,8}{3}\) => Sau phản ứng H2SO4 dư
\(m_{H_2SO_4}=\left(0,8-0,2.3\right).98=19,6\left(g\right)\)
b)\(n_{Al_2\left(SO_4\right)_3}=n_{Al_2O_3}=0,2\left(mol\right)\)
\(m_{Al_2\left(SO_4\right)_3}=0,2.342=68,4\left(g\right)\)
Bài 5:
Ta có: \(n_P=\dfrac{6,2}{31}=0,2\left(mol\right)\)
a, PT: \(4P+5O_2\underrightarrow{t^o}2P_2O_5\)
_____0,2__0,25__0,1 (mol)
b, VO2 = 0,25.22,4 = 5,6 (l)
c, PT: \(P_2O_5+3H_2O\rightarrow2H_3PO_4\)
______0,1______________0,2 (mol)
\(\Rightarrow m_{H_3PO_4}=0,2.98=19,6\left(g\right)\)
\(\Rightarrow C\%_{H_3PO_4}=\dfrac{19,6}{120}.100\%\approx16,33\text{ }\%\)
Bạn tham khảo nhé!
a, \(CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O\)
b, \(n_{CaCO_3}=\dfrac{10}{100}=0,1\left(mol\right)\Rightarrow n_{CO_2}=n_{CaCO_3}=0,1\left(mol\right)\)
\(\Rightarrow V_{CO_2}=0,1.22,4=2,24\left(l\right)\)
c, \(n_{HCl}=2n_{CaCO_3}=0,2\left(mol\right)\Rightarrow C_{M_{HCl}}=\dfrac{0,2}{0,1}=2\left(M\right)\)
d, \(m_{NaOH}=550.10\%=55\left(g\right)\Rightarrow n_{NaOH}=\dfrac{55}{40}=1,375\left(mol\right)\)
\(\Rightarrow\dfrac{n_{NaOH}}{n_{CO_2}}=\dfrac{1,375}{0,1}=13,75>2\)
→ Pư tạo muối trung hòa Na2CO3.
PT: \(CO_2+2NaOH\rightarrow Na_2CO_3+H_2O\)
\(n_{Na_2CO_3}=n_{CO_2}=0,1\left(mol\right)\Rightarrow m_{Na_2CO_3}=0,1.106=10,6\left(g\right)\)
a) \(n_{KMnO_4}=\dfrac{47,4}{158}=0,3\left(mol\right)\)
PTHH: 2KMnO4 --to--> K2MnO4+ MnO2 + O2
0,3------------------------------>0,15
=> VO2 = 0,15.22,4 = 3,36 (l)
b)
\(n_P=\dfrac{3,1}{31}=0,1\left(mol\right)\)
PTHH: 4P + 5O2 --to--> 2P2O5
Xét tỉ lệ: \(\dfrac{0,1}{4}< \dfrac{0,15}{5}\) => P hết, O2 dư
PTHH: 4P + 5O2 --to--> 2P2O5
0,1------------>0,05
P2O5 + 3H2O --> 2H3PO4
0,05-------------->0,1
=> \(C_{M\left(H_3PO_4\right)}=\dfrac{0,1}{0,08}=1,25M\)
a)2KMnO4→K2MnO4+MnO2+O2nKMnO4=47,4158=0,3molnO2=nKMnO42=0,15molVO2=0,15×22,4=3,36lb)O2+4Na→2Na2ONa2O+H2O→2NaOHnNa=9,223=0,4mol0,44<0,151⇒O2 dưnNa2O=nNa2=0,2molnNaOH=2nNa2O=0,4molCMNaOH=0,40,2=2Mc)nNaOH=0,1×2=0,2molVNaOH=nCM=0,20,5=0,4l=400mla)2KMnO4→K2MnO4+MnO2+O2nKMnO4=47,4158=0,3molnO2=nKMnO42=0,15molVO2=0,15×22,4=3,36lb)O2+4Na→2Na2ONa2O+H2O→2NaOHnNa=9,223=0,4mol0,44<0,151⇒O2 dưnNa2O=nNa2=0,2molnNaOH=2nNa2O=0,4molCMNaOH=0,40,2=2Mc)nNaOH=0,1×2=0,2molVNaOH=nCM=0,20,5=0,4l=400ml
Cách tiến hành : Cho thêm 300ml nước vào dung dịch A
\(nNa=\dfrac{6,9}{23}=0,3\left(mol\right)\)
\(4Na+O_2\underrightarrow{t^o}2Na_2O\)
4 1 2 (mol)
0,3 0,075 0,15
\(VO_2=0,075.22,4=1,68\left(l\right)\)
\(Na_2O+H_2O\rightarrow2NaO H\)
1 1 2 (mol)
0,15 0,15 0,3 (mol)
\(m_{NaOH}=0,3.40=12\left(g\right)\)
\(C\%_{ddA}=\dfrac{12.100}{180}=6,67\%\)