\(4P+5O_2\underrightarrow{t^o}2P_2O_5\)

Theo ĐLBTKL, ta có:

\(m_P+m_{O_2}=m_{P_2O_5}\\ \Rightarrow m_{O_2}=7,1-3,1=4g\)

Ta có : \(m_P+m_o=m_{p_2O_5}\)

hay  \(3,1+5O_2=7,1\)

       \(\Rightarrow m_{O_2}=7,1=3,1=4\left(gam\right)\)

 1)

a) photpho+oxi--->điphotpho pentaoxit

b)M photpho+M oxi---> M điphotpho pentaoxit

3.1 + M oxi--> 7.1

M oxi = 7.1 - 3.1 = 4g

mới làm xong bài 1 bấm xem thêm thấy bất ngờ lun nên thôi ko lm nx hihi

Sửa câu b: tính kl oxi đã dùng

\(a,4P+5O_2\xrightarrow{t^o}2P_2O_5\\ b,\text{Bảo toàn KL: }m_P+m_{O_2}=m_{P_2O_5}\\ \Rightarrow m_{O_2}=7,1-3,1=4(g)\)

`4P + 5O_2` $\xrightarrow{t^o}$ `2P_2 O_5`

`0,1`                        `0,05`               `(mol)`

`n_P=[3,1]/31=0,1(mol)`

`=>m_[P_2 O_5]=0,05.142=7,1(g)`

\(PTHH:4P+5O_2-^{t^o}>2P_2O_5\)

áp dụng ĐLBTKL ta có

\(m_P+m_{O_2}=m_{P_2O_5}\\ =>m_{O_2}=m_{P_2O_5}-m_P\\ =>m_{O_2}=7,1-3,1\\ =>m_{O_2}=4\left(g\right)\)

a) $\rm 4P + 5O_2 \xrightarrow{t^o} 2P_2O_5$
b) Áp dụng ĐLBTKL:
$\rm m_P + m_{O_2} = m_{P_2O_5}$
$\rm \Rightarrow m_{O_2} = m_{P_2O_5} - m_P = 7,1 - 3,1 = 4 (g)$

\(n_P=\dfrac{3,1}{31}=0,1\left(mol\right)\)

PTHH: 4P + 5O₂ --t^o--> 2P₂O₅

_____0,1-->0,125------>0,05_______(mol)

=> \(\left\{{}\begin{matrix}m_{O_2}=0,125.32=4\left(g\right)\\m_{P_2O_5}=0,05.142=7,1\left(g\right)\end{matrix}\right.\)

Ta có: \(n_P=\dfrac{3,1}{31}=0,1\left(mol\right)\)

PT: \(4P+5O_2\underrightarrow{t^o}2P_2O_5\)

___0,1__0,125___0,05 (mol)

\(\Rightarrow m_{O_2}=0,125.32=4\left(g\right)\)

\(m_{P_2O_5}=0,05.142=7,1\left(g\right)\)

Bạn tham khảo nhé!

a) 4P + 5O₂ --t^o--> 2P₂O₅

b) \(n_P=\dfrac{3,1}{31}=0,1\left(mol\right)\)

PTHH: 4P + 5O₂ --t^o--> 2P₂O₅

_____0,1-->0,125---->0,05

=> m_P2O5 = 0,05.142 = 7,1 (g)

c) V_O2 = 0,125.22,4 = 2,8(l)

a) 4P + 5O₂ --t^o--> 2P₂O₅

b)0,1mol

c, 2,8l

\(n_P=\dfrac{6,2}{31}=0,2\left(mol\right)\\ n_{O_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\\ PTHH:4P+5O_2\underrightarrow{t^o}2P_2O_5\\ LTL:\dfrac{0,2}{4}< \dfrac{0,4}{5}\Rightarrow O_2dư\)

\(n_{O_2\left(pư\right)}=\dfrac{5}{4}n_P=\dfrac{5}{4}.0,2=0,25\left(mol\right)\\ n_{O_2\left(dư\right)}=0,4-0,25=0,15\left(mol\right)\)

\(n_{P_2O_5\left(lt\right)}=\dfrac{1}{2}n_P=\dfrac{1}{2}.0,2=0,1\left(mol\right)\\ m_{P_2O_5\left(lt\right)}=0,1.142=14,2\left(g\right)\\ m_{P_2O_5\left(tt\right)}=0,1.142.80\%=11,36\left(g\right)\)