Cho 10 lít dung dịch HNO3 63% (D=1,4g/ml) phản ứng với xenlulozơ dư thu được

Đáp án D.

$$\begin{gathered} {\left[{\text{C}}_6{\text{H}}_7{\text{O}}_2{\left(\text{OH}\right)}_3\right]}_n+3n{\text{HNO}}_3\stackrel{{\text{H}}_2{\text{SO}}_4}{\to }{\left[{\text{C}}_6{\text{H}}_7{\text{O}}_2{\left({\text{ONO}}_2\right)}_3\right]}_n+3n{\text{H}}_2\text{O} \\ {n}_{{\text{HNO}}_3\left(\text{ ban đầu }\right)}=\frac{V_{dd}.d.C\%}{63}=\frac{\mathrm{10.1000.1},4.63\%}{63}=140 \text{\hspace{0.17em}\hspace{0.17em}mol} \\ {n}_{{\text{HNO}}_3(\text{ (phản ứng) }}=n_{{\text{HNO}}_3\left(\text{ ban đầu }\right)}.H\%=126 \text{\hspace{0.17em}\hspace{0.17em}mol} \\ {n}_{\text{xenlulozô trinitrat }}=\frac{1}{3n}{n}_{{\text{HNO}}_3\left(\text{ phản ứng }\right)}=\frac{42}{n} \text{\hspace{0.17em}\hspace{0.17em}mol} \\ {m}_{\text{xenlulozô trinitrat }}=\frac{0,042}{n}\cdot 297n=12474\text{\hspace{0.17em}\hspace{0.17em}gam}=12,474( \text{kg}) \end{gathered}$$