\(a,n_{Na}=\dfrac{2,3}{23}=0,1\left(mol\right);n_{O_2}=\dfrac{0,896}{22,4}=0,04\left(mol\right)\\ a,PTHH:4Na+O_2\rightarrow2Na_2O\\ Vì:\dfrac{0,1}{4}< \dfrac{0,04}{1}\Rightarrow O_2dư\\ n_{O_2\left(dư\right)}=0,04-\dfrac{0,1}{4}=0,015\left(mol\right)\\ \Rightarrow m_{O_2\left(dư\right)}=0,015.32=0,48\left(g\right)\\ b,C1:n_{Na_2O}=\dfrac{2}{4}.n_{Na}=\dfrac{2}{4}.0,1=0,05\left(mol\right)\\ \Rightarrow m_{Na_2O}=0,05.62=3,1\left(g\right)\\ C2:ĐLBTKL:m_{Na_2O}=m_{Na}+m_{O_2\left(bđ\right)}-m_{O_2\left(dư\right)}=2,3+0,04.32-0,48=3,1\left(g\right)\)
a. \(n_{Na}=\dfrac{2,3}{23}=0,1\left(mol\right)\)
\(n_{O_2}=\dfrac{0,896}{22,4}=0,04\left(mol\right)\)
PTHH : 4Na + O2 -to> 2Na2O
0,1 0,025 0,05
Xét tỉ lệ : \(\dfrac{0,1}{4}< \dfrac{0,04}{1}\) => Na đủ , O2 dư
\(m_{O_2\left(dư\right)}=\left(0,04-0,025\right).32=0,48\left(g\right)\)
b. Cách 1 : \(m_{Na_2O}=0,05.62=3,1\left(g\right)\)
Cách 2 : \(m_{Na}=0,1.23=2,3\left(g\right)\)
\(m_{O_2}=0,025.32=0,8\left(g\right)\)
Theo ĐLBTKL:
\(m_{Na}+m_{O_2}=m_{Na_2O}\\ \Rightarrow2,3+0,8=3,1\left(g\right)\)
a,pthh: \(4Na+O_2\underrightarrow{t^o}2Na_2O\)
\(nNa=\dfrac{2,3}{23}=0,1\left(mol\right)\)
\(nO_2=0,896:22,4=0,04\left(mol\right)\)
Xét tỉ lệ : \(\dfrac{nNa}{4}< \dfrac{nO_2}{1}\) ( \(\dfrac{0,1}{4}< \dfrac{0,04}{1}\)) => Oxi dư
\(\Rightarrow nO_{2\left(dư\right)}=0,04-0,025=0,015\left(mol\right)\)
\(nO_{2\left(dư\right)}=0,015.32=0,48\left(g\right)\)
b, do oxi dư nên lấy số mol của natri làm chuẩn.
\(nNa_2O=0,05\left(mol\right)\)
\(mNa_2O=0,05.\left(23.2+16\right)=3,1\left(gam\right)\)
c2: \(mO_{2\left(đủ\right)}=\) 0,8(gam)
BTKL : mNa+mO2=mNa2O
<=> 2,3+ 0,8= mNa2O
=>mNa2O= 3,1 (gam)
