\(nMg=\frac{2.4}{24}=0,1\left(mol\right)\) ; \(nO_2=\frac{3,2}{32}=0,1\left(mol\right)\)
PTHH
2Mg + O2 ----> 2MgO
2.........1.................2
Ta có : \(\frac{0,1}{2}< \frac{0,1}{1}\) => Mg dư
\(nMgdư=0,1-\frac{0,1}{2}=0,05\left(mol\right)\)\(\Rightarrow m=0,05\cdot24=1,2\left(g\right)\)
\(nMg=\frac{1}{2}nO_2=0,05\left(mol\right)\)
\(mMgO=0,05\cdot40=2\left(g\right)\)
\(n_{Mg}=\frac{2,4}{24}=0,1\left(mol\right)\)
\(n_{O2}=\frac{3,2}{32}=0,1\left(mol\right)\)
\(2Mg+O_2\rightarrow2MgO\)
Xét tỷ số: \(\frac{0,1}{2}< \frac{0,1}{1}\)
Mg hết, O2 dư \(\Rightarrow\) Tính theo nMg
\(n_{O2}=\frac{1}{2}n_{Mg}=0,05\left(mol\right)\)
\(n_{O2_{dư}}=0,1-0,05=0,05\left(mol\right)\)
\(\Rightarrow m_{O2_{dư}}=0,05.32=1,6\left(g\right)\)
b. \(n_{MgO2}=n_{Mg}=0,1\left(mol\right)\)
\(\Rightarrow m_{MgO}=0,1.37=3,7\left(g\right)\)
\(2Mg+O_2\overset{t^0}{\rightarrow}2MgO\)
2:1(tỉ lệ)
a)\(n_{Mg}=\frac{2.4}{24}=0.1\left(mol\right)\)
\(n_{O_2}=\frac{3.2}{32}=0.1\left(mol\right)\)
So sánh: \(\frac{0.1}{2}< \frac{0.1}{1}\Rightarrow O_2\) dư, bài toán tính theo Mg
\(\Rightarrow n_{O_2\left(pư\right)}=\frac{1}{2}\cdot n_{Mg}=\frac{1}{2}\cdot0.1=0.05\left(mol\right)\)
\(\Rightarrow n_{O_2\left(dư\right)}=n_{O_2}-m_{O_2\left(pư\right)}=0.1-0.05=0.05\left(mol\right)\)
\(\Rightarrow m_{O_2\left(dư\right)}=0.05\cdot32=1.8\left(g\right)\)
b)\(n_{MgO}=\frac{2}{2}\cdot n_{Mg}=\frac{2}{2}\cdot0.1=0.1\left(mol\right)\)
\(\Rightarrow m_{MgO}=0.1\cdot40=4\left(g\right)\)
Chỗ \(m_{O_2\left(dư\right)}=1.6\left(g\right)\), mình tính nhầm!!!
