Đáp án + Giải thích các bước giải:
a, $\text{$m_{CO_{2}}$ = 4,4 (g) $\Rightarrow$ $n_{CO_{2}}$ = $\dfrac{m_{CO_{2}}}{M_{CO_{2}}}$ = $\dfrac{4,4}{12 + 16.2}$ = $\dfrac{4,4}{44}$ = 0,1 (mol)}$
b, $\text{$V_{N_{2}}$ = 4,48 (l) $\Rightarrow$ $n_{N_{2}}$ = $\dfrac{V_{N_{2}}}{22,4}$ = $\dfrac{4,48}{22,4}$ = 0,2 (mol)}$
c, $\text{$V_{NH_{3}}$ = 3,36 (l) $\Rightarrow$ $n_{NH_{3}}$ = $\dfrac{V_{NH_{3}}}{22,4}$ = $\dfrac{3,36}{22,4}$ = 0,15 (mol)}$
$\text{$\Longrightarrow$ $m_{NH_{3}}$ = $n_{NH_{3}}$ . $M_{NH_{3}}$ = 0,15 . (12 + 1.3) = 0,15 . 15 = 2,25 (g)}$
d, $\text{C% = $\dfrac{m_{ct}}{m_{dd}}$ . 100 = $\dfrac{20}{180}$ . 100 $\approx$ 11,11%}$
e, $\text{$C_{M_{HNO_{3}}}$ = $\dfrac{n_{HNO_{3}}}{V_{HNO_{3}}}$ ⇔ 2 = $\dfrac{n_{HNO_{3}}}{\dfrac{500}{1000}}$ ⇔ 2 = $\dfrac{n_{HNO_{3}}}{0,5}$ ⇔ $n_{HNO_{3}}$ = 1 (mol)}$
$\text{⇒ $m_{HNO_{3}}$ = $n_{HNO_{3}}$ . $M_{HNO_{3}}$ = 1 . (1 + 14 + 16.3) = 1 . 63 = 63 (g)}$
CHÚC BẠN HỌC TỐT:333
