Question

Answer 1

a, \(n_{P_2O_5}=\dfrac{42,6}{142}=0,3\left(mol\right)\)

\(n_{H_2O}=\dfrac{108}{18}=6\left(mol\right)\)

PTHH: P₂O₅ + 3H₂O ---> 2H₃PO₄

LTL: \(0,3< \dfrac{6}{3}\rightarrow\)H₂O dư

Theo pt: \(\left\{{}\begin{matrix}n_{H_3PO_4}=2n_{P_2O_5}=2.0,3=0,6\left(mol\right)\\n_{H_2O\left(pư\right)}=3n_{P_2O_5}=0,3.3=0,9\left(mol\right)\end{matrix}\right.\)

=> Số ptử H₃PO₄: 0,6.6.10²³ = 3,6.10²³ (ptử)

=> Số ptử H₂O (dư): (6 - 0,9).6²³ = 30,6.10²³ (ptử)

b, Do m_Y = m_X

=> m_{O2 (sinh ra từ KClO3)} = m_{O2 (pư vs Cu)}

=> n_{O2 (sinh ra từ KClO3)} = n_{O2 (pư vs Cu)}

PTHH:

2KClO₃ --t^o--> 2KCl + 3O₂

a --------------------------> 1,5a

2Cu + O₂ --t^o--> 2CuO

 3a <--- 1,5a

Mà n_Cu = b (mol)

=> 3a = b