Đáp án:
Giải thích các bước giải:
\(\text{Bài 6:}\\ a,\ n_{NaCl}=0,5\times 1=0,5\ mol.\\ ⇒m_{NaCl}=0,5\times 58,5=29,25\ g.\\ b,\ n_{CaCl_2}=0,25\times 0,1=0,025\ mol.\\ ⇒m_{CaCl_2}=0,025\times 111=2,775\ g.\\ \text{Bài 7:}\\ a,\ C\%_{KCl}=\dfrac{20}{600}\times 100\%=3,33\%\\ b,\ C\%_{K_2SO_4}=\dfrac{75}{1500}\times 100\%=5\%\\ \text{Bài 8:}\\ a,\ m_{Fe}=0,2\times 56=11,2\ g.\\ m_{Cu}=0,5\times 64=32\ g.\\ ⇒m_{\text{chất rắn}}=m_{Fe}+m_{Cu}=11,2+32=43,2\ g.\\ b,\ n_{CO_2}=\dfrac{33,6}{22,4}=1,5\ mol.\\ ⇒m_{CO_2}=1,5\times 44=66\ g.\\ n_{CO}=\dfrac{11,2}{22,4}=0,5\ mol.\\ ⇒m_{CO}=0,5\times 28=14\ g.\\ n_{N_2}=\dfrac{5,6}{22,4}=0,25\ mol.\\ ⇒m_{N_2}=0,25\times 28=7\ g.\\ ⇒m_{\text{chất rắn}}=m_{CO_2}+m_{CO}+m_{N_2}=66+14+7=87\ g.\\ \text{Bài 9:}\\ a,\ n_{\text{hh khí}}=(0,75+0,5+0,25)\times 22,4=33,6\ lít.\\ b,\ n_{O_2}=\dfrac{6,4}{32}=0,2\ mol.\\ n_{N_2}=\dfrac{22,4}{28}=0,8\ mol.\\ ⇒V_{\text{hh khí}}=(0,2+0,8)times 22,4=22,4\ lít.\)
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