Câu hỏi của Vyy Wibu

Question

để thủy phân hoàn toàn 4,45kg một loại chất béo(C17H35COO03C3H5 cới dung dịch NaOH vừa thu được glixerol và muối.

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๖ۣۜDũ๖ۣۜN๖ۣۜG · Accepted Answer

$n_{\left(C_{17}H_{35}COO\right)_3C_3H_5\left(bđ\right)}=\dfrac{4,45.10^3}{890}=5\left(mol\right)$

=> $n_{\left(C_{17}H_{35}COO\right)_3C_3H_5\left(pư\right)}=\dfrac{5.80}{100}=4\left(mol\right)$

=> $n_{C_{17}H_{35}COONa}=3.n_{\left(C_{17}H_{35}COO\right)_3C_3H_5\left(pư\right)}=12\left(mol\right)$

=> $m_{C_{17}H_{35}COONa}=12.306=3672\left(g\right)$

=> $m_{xà.phòng}=\dfrac{3672.100}{62}=\dfrac{183600}{31}\left(g\right)$

Câu trả lời:

$n_{\left(C_{17}H_{35}COO\right)_3C_3H_5\left(bđ\right)}=\dfrac{4,45.10^3}{890}=5\left(mol\right)$

=> $n_{\left(C_{17}H_{35}COO\right)_3C_3H_5\left(pư\right)}=\dfrac{5.80}{100}=4\left(mol\right)$

=> $n_{C_{17}H_{35}COONa}=3.n_{\left(C_{17}H_{35}COO\right)_3C_3H_5\left(pư\right)}=12\left(mol\right)$

=> $m_{C_{17}H_{35}COONa}=12.306=3672\left(g\right)$

=> $m_{xà.phòng}=\dfrac{3672.100}{62}=\dfrac{183600}{31}\left(g\right)$