Tính phần trăm về khối lượng nguyên tố: a, NaCl b, Fe2O3 c, Cu0 d, NaNO3 14/07/2021 Bởi Eloise Tính phần trăm về khối lượng nguyên tố: a, NaCl b, Fe2O3 c, Cu0 d, NaNO3
a, $M_{NaCl}$ = 23 + 35,5 = 58,5 %$m_{Na}$ = $\frac{23}{58,5}$ ~ 39,31% %$m_{Cl}$ = 100% – 39,31% = 60,68% b, $M_{Fe_{2}O_{3}}$ = 56.2 + 16.3 = 160 %$m_{Fe}$ = $\frac{112}{160}$ = 70% %$m_{O}$ = 100% – 70% = 30% c, $M_{CuO}$ = 64.1 + 16.1 = 80 %$m_{Cu}$ = $\frac{64}{80}$ = 80% %$m_{O}$ = 100% – 80% = 20% d, $M_{NaNO_{3}}$ = 23 + 14 + 3.16 = 85 %$m_{Na}$ = $\frac{23}{85}$ = 27,05% %$m_{N}$ = $\frac{14}{85}$ ~ 16,47% %$m_{O}$ = 100% – 27,05% – 16,47% = 56,48% Bình luận
Đáp án: `a)` `M_(NaCl)=23+35,5=58,5 \ \ (g//mol)` `to` `%m_(Na)=23/(58,5) . 100% = 39,3%` `%m_(Cl)=100% – 39,3% = 60,7%` `b)` `M_(Fe_2O_3)=56.2+16.3=160 \ \ (g//mol)` `to` `%m_(Fe)=(56.2)/160 . 100% = 70%` `%m_O=100%-70%=30%` `c)` `M_(CuO)=64+16=80 \ \ (g//mol)` `to` `%m_(Cu)=64/80 . 100%=80%` `%m_O=100%-80%=20%` `d)` `M_(NaNO_3)=23+14+16.3=85 \ \ (g//mol)` `to` `%m_(Na)=23/85 . 100%=27,1%` `%m_N=14/85 . 100%=16,5%` `%m_O=100%-27,1%-16,5%=56,4%` Bình luận
a, $M_{NaCl}$ = 23 + 35,5 = 58,5
%$m_{Na}$ = $\frac{23}{58,5}$ ~ 39,31%
%$m_{Cl}$ = 100% – 39,31% = 60,68%
b, $M_{Fe_{2}O_{3}}$ = 56.2 + 16.3 = 160
%$m_{Fe}$ = $\frac{112}{160}$ = 70%
%$m_{O}$ = 100% – 70% = 30%
c, $M_{CuO}$ = 64.1 + 16.1 = 80
%$m_{Cu}$ = $\frac{64}{80}$ = 80%
%$m_{O}$ = 100% – 80% = 20%
d, $M_{NaNO_{3}}$ = 23 + 14 + 3.16 = 85
%$m_{Na}$ = $\frac{23}{85}$ = 27,05%
%$m_{N}$ = $\frac{14}{85}$ ~ 16,47%
%$m_{O}$ = 100% – 27,05% – 16,47% = 56,48%
Đáp án:
`a)`
`M_(NaCl)=23+35,5=58,5 \ \ (g//mol)`
`to`
`%m_(Na)=23/(58,5) . 100% = 39,3%`
`%m_(Cl)=100% – 39,3% = 60,7%`
`b)`
`M_(Fe_2O_3)=56.2+16.3=160 \ \ (g//mol)`
`to`
`%m_(Fe)=(56.2)/160 . 100% = 70%`
`%m_O=100%-70%=30%`
`c)`
`M_(CuO)=64+16=80 \ \ (g//mol)`
`to`
`%m_(Cu)=64/80 . 100%=80%`
`%m_O=100%-80%=20%`
`d)`
`M_(NaNO_3)=23+14+16.3=85 \ \ (g//mol)`
`to`
`%m_(Na)=23/85 . 100%=27,1%`
`%m_N=14/85 . 100%=16,5%`
`%m_O=100%-27,1%-16,5%=56,4%`