Question

Tính khối lượng của:
1) 1,2 mol Cu 
2) 1,25 mol NaCl 
3) 7,2 .10²³ phân tử C₆H₁₂O₆
4) 3,6 mol O₂ 

5) 1,2 . 10²³ phân tử khí O₂

6) 26,881 khí N₂

7) 11,21 khí CO₂

8) 31,36 l khí H₂

 

 

Answer 1

\(1.m_{Cu}=1,2.64=76,8\left(g\right)\\ 2.m_{NaCl}=1,25.58,5=73,125\\ 3.n_{C_6H_{12}O_6}=\dfrac{7,2.10^{23}}{6.10^{23}}=1,2\left(mol\right)\\ \Rightarrow m_{C_6H_{12}O_6}=1,2.180=216\left(g\right)\\ 4.n_{O_2}=3,6.32=115,2\left(g\right)\\ 5.n_{O_2}=\dfrac{1,2.10^{23}}{6.10^{23}}=0,2\left(mol\right)\\ \Rightarrow m_{O_2}=0,2.32=6,4\left(g\right)\\ 6.n_{N_2}=\dfrac{26,88}{22,4}=1,2\left(mol\right)\\ \Rightarrow m_{N_2}=1,2.28=33,6\left(g\right)\\ 7.n_{CO_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\\ \Rightarrow m_{CO_2}=0,5.44=22\left(g\right)\\ 8.n_{H_2}=\dfrac{31,36}{22,4}=1,4\left(mol\right)\\ \Rightarrow m_{H_2}=1,4.2=2,8\left(g\right)\)

Answer 2

\(1,m_{Cu}=1,2\cdot64=76,8\left(g\right)\\ 2,m_{NaCl}=1,25\cdot58,5=73,125\left(g\right)\\ 3,n_{C_6H_{12}O_6}=\dfrac{7,2\cdot10^{-23}}{6\cdot10^{-23}}=1,2\left(mol\right)\\ \Rightarrow m_{C_6H_{12}O_6}=1,2\cdot180=216\left(g\right)\\ 4,m_{O_2}=3,6\cdot32=115,2\left(g\right)\\ 5,n_{O_2}=\dfrac{1,2\cdot10^{-23}}{6\cdot10^{-23}}=0,2\left(mol\right)\\ \Rightarrow m_{O_2}=0,2\cdot32=6,4\left(g\right)\\ 6,n_{N_2}=\dfrac{26,88}{22,4}=1,2\left(mol\right)\\ \Rightarrow m_{N_2}=1,2\cdot28=33,6\left(g\right)\\ 7,n_{CO_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\\ \Rightarrow m_{CO_2}=0,5\cdot44=22\left(g\right)\\ 8,n_{H_2}=\dfrac{31,36}{22,4}=1,4\left(mol\right)\\ \Rightarrow m_{H_2}=1,4\cdot2=2,8\left(g\right)\)