Câu hỏi của Nguyễn Vũ Ánh Dương

Question

Bài 1: Cho sơ đồ Cu   +   $O_2$   ---->CuOa, Tính khối lượng CuO sinh ra khi có 2,56g Cu tham gi phản ứng.b, Tính khối lượng Cu cần dùng để điều chế 4g CuO.c, Tính khối lượng Cu và thể tích khí oxi...

𝓓𝓾𝔂 𝓐𝓷𝓱 · Accepted Answer

Bài 2:PTHH: $Zn+2HCl\rightarrow ZnCl_2+H_2$a+b) Ta có: $n_{Zn}=\dfrac{16,25}{65}=0,25\left(mol\right)$$\Rightarrow\left\{{}\begin{matrix}n_{HCl}=0,5\left(mol\right)\n_{H_2}=0,25\left(mol\right)\end{matrix}\right.$ $\Rightarrow\left\{{}\begin{matrix}m_{HCl}=0,5\cdot36,5=18,25\left(g\right)\V_{H_2}=0,25\cdot22,4=5,6\left(l\right)\end{matrix}\right.$c) +) Cách 1: Theo PTHH: $n_{ZnCl_2}=n_{Zn}=0,25\left(mol\right)$ $\Rightarrow n_{ZnCl_2}=0,25\cdot136=34\left(g\right)$+) Cách 2: Ta có: $m_{H_2}=0,25\cdot2=0,5\left(g\right)$Bảo toàn khối lượng: $m_{ZnCl_2}=m_{Zn}+m_{HCl}-m_{H_2}=16,25+18,25-0,5=34\left(g\right)$Câu trả lời: Bài 2:PTHH: $Zn+2HCl\rightarrow ZnCl_2+H_2$a+b) Ta có: $n_{Zn}=\dfrac{16,25}{65}=0,25\left(mol\right)$$\Rightarrow\left\{{}\begin{matrix}n_{HCl}=0,5\left(mol\right)\n_{H_2}=0,25\left(mol\right)\end{matrix}\right.$ $\Rightarrow\left\{{}\begin{matrix}m_{HCl}=0,5\cdot36,5=18,25\left(g\right)\V_{H_2}=0,25\cdot22,4=5,6\left(l\right)\end{matrix}\right.$c) +) Cách 1: Theo PTHH: $n_{ZnCl_2}=n_{Zn}=0,25\left(mol\right)$ $\Rightarrow n_{ZnCl_2}=0,25\cdot136=34\left(g\right)$+) Cách 2: Ta có: $m_{H_2}=0,25\cdot2=0,5\left(g\right)$Bảo toàn khối lượng: $m_{ZnCl_2}=m_{Zn}+m_{HCl}-m_{H_2}=16,25+18,25-0,5=34\left(g\right)$

Lê Ng Hải Anh · Answer

Bài 1:PT: $2Cu+O_2\underrightarrow{t^o}2CuO$a, Ta có: $n_{Cu}=\dfrac{2,56}{64}=0,04\left(mol\right)$Theo PT: $n_{Cu}=n_{CuO}=0,04\left(mol\right)$$\Rightarrow m_{CuO}=0,04.80=3,2\left(g\right)$b, Ta có: $n_{CuO}=\dfrac{4}{80}=0,05\left(mol\right)$Theo PT: $n_{Cu}=n_{CuO}=0,05\left(mol\right)$$\Rightarrow m_{Cu}=0,05.64=3,2\left(g\right)$c, Ta có: $n_{CuO}=\dfrac{24}{80}=0,3\left(mol\right)$Theo PT: $\left\{{}\begin{matrix}n_{Cu}=n_{CuO}=0,3\left(mol\right)\n_{O_2}=\dfrac{1}{2}n_{CuO}=0,15\left(mol\right)\end{matrix}\right.$$\Rightarrow m_{Cu}=0,3.64=19,2\left(g\right)$$V_{O_2}=0,15.22,4=3,36\left(l\right)$Bài 2: PT: $Zn+2HCl\rightarrow ZnCl_2+H_2$a, Ta có: $m_{Zn}=\dfrac{16,25}{65}=0,25\left(mol\right)$Theo PT: $n_{HCl}=2n_{Zn}=0,5\left(mol\right)$$\Rightarrow m_{HCl}=0,5.36,5=18,25\left(g\right)$b, Theo PT: $n_{H_2}=n_{Zn}=0,25\left(mol\right)$$\Rightarrow V_{H_2}=0,25.22,4=5,6\left(l\right)$c, Cách 1: Theo PT: $n_{ZnCl_2}=n_{Zn}=0,25\left(mol\right)$$\Rightarrow m_{ZnCl_2}=0,25.136=34\left(g\right)$Cách 2:Theo ĐLBT KL, có: mZn + mHCl = mZnCl2 + mH2⇒ mZnCl2 = mZn + mHCl - mH2 = 16,25 + 18,25 - 0,25.2 = 34 (g)Bạn tham khảo nhé! Câu trả lời: Bài 1:PT: $2Cu+O_2\underrightarrow{t^o}2CuO$a, Ta có: $n_{Cu}=\dfrac{2,56}{64}=0,04\left(mol\right)$Theo PT: $n_{Cu}=n_{CuO}=0,04\left(mol\right)$$\Rightarrow m_{CuO}=0,04.80=3,2\left(g\right)$b, Ta có: $n_{CuO}=\dfrac{4}{80}=0,05\left(mol\right)$Theo PT: $n_{Cu}=n_{CuO}=0,05\left(mol\right)$$\Rightarrow m_{Cu}=0,05.64=3,2\left(g\right)$c, Ta có: $n_{CuO}=\dfrac{24}{80}=0,3\left(mol\right)$Theo PT: $\left\{{}\begin{matrix}n_{Cu}=n_{CuO}=0,3\left(mol\right)\n_{O_2}=\dfrac{1}{2}n_{CuO}=0,15\left(mol\right)\end{matrix}\right.$$\Rightarrow m_{Cu}=0,3.64=19,2\left(g\right)$$V_{O_2}=0,15.22,4=3,36\left(l\right)$Bài 2: PT: $Zn+2HCl\rightarrow ZnCl_2+H_2$a, Ta có: $m_{Zn}=\dfrac{16,25}{65}=0,25\left(mol\right)$Theo PT: $n_{HCl}=2n_{Zn}=0,5\left(mol\right)$$\Rightarrow m_{HCl}=0,5.36,5=18,25\left(g\right)$b, Theo PT: $n_{H_2}=n_{Zn}=0,25\left(mol\right)$$\Rightarrow V_{H_2}=0,25.22,4=5,6\left(l\right)$c, Cách 1: Theo PT: $n_{ZnCl_2}=n_{Zn}=0,25\left(mol\right)$$\Rightarrow m_{ZnCl_2}=0,25.136=34\left(g\right)$Cách 2:Theo ĐLBT KL, có: mZn + mHCl = mZnCl2 + mH2⇒ mZnCl2 = mZn + mHCl - mH2 = 16,25 + 18,25 - 0,25.2 = 34 (g)Bạn tham khảo nhé!

𝓓𝓾𝔂 𝓐𝓷𝓱 · Answer

Bài 1: PTHH: $2Cu+O_2\underrightarrow{t^o}2CuO$a) Ta có: $n_{Cu}=n_{CuO}=\dfrac{2,56}{64}=0,04\left(mol\right)$ $\Rightarrow m_{CuO}=0,04\cdot80=3,2\left(g\right)$b) Ta có: $n_{Cu}=n_{CuO}=\dfrac{4}{80}=0,05\left(mol\right)$ $\Rightarrow m_{Cu}=0,05\cdot64=3,2\left(g\right)$c) Ta có: $n_{CuO}=\dfrac{24}{80}=0,3\left(mol\right)$$\Rightarrow\left\{{}\begin{matrix}n_{Cu}=0,3\left(mol\right)\n_{O_2}=0,15\left(mol\right)\end{matrix}\right.$ $\Rightarrow\left\{{}\begin{matrix}m_{Cu}=0,3\cdot64=19,2\left(g\right)\V_{O_2}=0,15\cdot22,4=3,36\left(l\right)\end{matrix}\right.$Câu trả lời: Bài 1: PTHH: $2Cu+O_2\underrightarrow{t^o}2CuO$a) Ta có: $n_{Cu}=n_{CuO}=\dfrac{2,56}{64}=0,04\left(mol\right)$ $\Rightarrow m_{CuO}=0,04\cdot80=3,2\left(g\right)$b) Ta có: $n_{Cu}=n_{CuO}=\dfrac{4}{80}=0,05\left(mol\right)$ $\Rightarrow m_{Cu}=0,05\cdot64=3,2\left(g\right)$c) Ta có: $n_{CuO}=\dfrac{24}{80}=0,3\left(mol\right)$$\Rightarrow\left\{{}\begin{matrix}n_{Cu}=0,3\left(mol\right)\n_{O_2}=0,15\left(mol\right)\end{matrix}\right.$ $\Rightarrow\left\{{}\begin{matrix}m_{Cu}=0,3\cdot64=19,2\left(g\right)\V_{O_2}=0,15\cdot22,4=3,36\left(l\right)\end{matrix}\right.$

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