10 g mixture of two gases A_2 (mol. wt. = 20 | vedclass

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(A) Let the initial mass of $A_2$ be $x \ g$ and $B_2$ be $y \ g$. Given: $x + y = 10 \ g$ (Equation $1$). For first-order kinetics,the remaining amou

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$4$

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(A) Let the initial mass of $A_2$ be $x \ g$ and $B_2$ be $y \ g$. Given: $x + y = 10 \ g$ (Equation $1$). For first-order kinetics,the remaining amount is given by $N = N_0 	imes (1/2)^n$,where $n$ is the number of half-lives. For $A_2$: Half-life $t_{1/2} = 2 \ hours$. In $6 \ hours$,$n = 6/2 = 3$ half-lives. Remaining $A_2 = x 	imes (1/2)^3 = x/8$. For $B_2$: Half-life $t_{1/2} = 3 \ hours$. In $6 \ hours$,$n = 6/3 = 2$ half-lives. Remaining $B_2 = y 	imes (1/2)^2 = y/4$. Total remaining mass after $6 \ hours = x/8 + y/4 = 2 \ g$ (Equation $2$). From Equation $1$,$y = 10 - x$. Substitute into Equation $2$: $x/8 + (10 - x)/4 = 2$. Multiply by $8$: $x + 2(10 - x) = 16$. $x + 20 - 2x = 16$. $-x = -4$,so $x = 4 \ g$.

Time (minute)	$0$	$20$	$75$	$120$	$\infty$
Volume of $0.05 \, M$ $NaOH$ ($V_t$ in $mL$)	$24.40$	$25.82$	$29.35$	$31.75$	$47.50$

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