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Câu hỏi: Cho 10 lít dung dịch $\mathrm{HNO}_{3} 63 \%$ (D=1,4g/ml) phản ứng với xenlulozơ dư thu được m kg thuốc súng không khói (xenlulozo trinitrat), biết hiệu suất phản ứng đạt $90 \%$. Giá trị m gần nhất với
A. $7,5.$
B. $6,5.$
C. $9,5.$
D. $12,5.$
A. $7,5.$
B. $6,5.$
C. $9,5.$
D. $12,5.$
${{\left[ {{\text{C}}_{6}}{{\text{H}}_{7}}{{\text{O}}_{2}}{{(\text{OH})}_{3}} \right]}_{n}}+3n\text{HN}{{\text{O}}_{3}}\xrightarrow{{{\text{H}}_{2}}\text{S}{{\text{O}}_{4}}}{{\left[ {{\text{C}}_{6}}{{\text{H}}_{7}}{{\text{O}}_{2}}{{\left( \text{ON}{{\text{O}}_{2}} \right)}_{3}} \right]}_{n}}+3n{{\text{H}}_{2}}\text{O}$
${{n}_{\text{HN}{{\text{O}}_{3}}(\text{ ban đầu })}}=\dfrac{{{V}_{dd}}.d.C\%}{63}=\dfrac{10.1000.1,4.63\%}{63}=140~ \text{mol}$
${{n}_{\text{HN}{{\text{O}}_{3}}(\text{ (phản ứng) }}}={{n}_{\text{HN}{{\text{O}}_{3}}(\text{ ban đầu })}}.H\%=126~ \text{mol}$
${{n}_{\text{xenluloz }\!\!\hat{\mathrm{o}}\!\!\text{ trinitrat }}}=\dfrac{1}{3n}{{n}_{\text{HN}{{\text{O}}_{3}}(\text{ phản ứng })}}=\dfrac{42}{n}~ \text{mol}$
${{m}_{\text{xenluloz }\!\!\hat{\mathrm{o}}\!\!\text{ trinitrat }}}=\dfrac{0,042}{n}\cdot 297n=12474 \text{gam}=12,474 (~\text{kg})$
${{n}_{\text{HN}{{\text{O}}_{3}}(\text{ ban đầu })}}=\dfrac{{{V}_{dd}}.d.C\%}{63}=\dfrac{10.1000.1,4.63\%}{63}=140~ \text{mol}$
${{n}_{\text{HN}{{\text{O}}_{3}}(\text{ (phản ứng) }}}={{n}_{\text{HN}{{\text{O}}_{3}}(\text{ ban đầu })}}.H\%=126~ \text{mol}$
${{n}_{\text{xenluloz }\!\!\hat{\mathrm{o}}\!\!\text{ trinitrat }}}=\dfrac{1}{3n}{{n}_{\text{HN}{{\text{O}}_{3}}(\text{ phản ứng })}}=\dfrac{42}{n}~ \text{mol}$
${{m}_{\text{xenluloz }\!\!\hat{\mathrm{o}}\!\!\text{ trinitrat }}}=\dfrac{0,042}{n}\cdot 297n=12474 \text{gam}=12,474 (~\text{kg})$
Đáp án D.