Question

cho 12 gam hon hop X (Fe, Cu) co ti le mol 1:1. Tinh khoi luong moi kim loai

Answer 1

Ta có: \(n_{Fe}=n_{Cu}=x\left(mol\right)\)

Ta có: \(m_X=m_{Fe}+m_{Cu}\)

\(\Leftrightarrow12=56x+64x\)

\(\Leftrightarrow12=120x\)

\(\Leftrightarrow x=0,1\left(mol\right)\)

Vậy \(m_{Fe}=0,1\times56=5,6\left(g\right)\)

\(m_{Cu}=0,1\times64=6,4\left(g\right)\)

Answer 2

Đặt:

nFe=nCu= x mol

mX = 56x + 64x = 12 g

=> x = 0.1

mFe= 5.6g

mCu= 6.4g

Answer 3

Gọi x (mol) là n_Cu và n_Fe \(\left(n_{Cu}=n_{Fe}\right)\)

Ta có mX= \(56x+64x=12\left(g\right)\)

⇔ \(120x=12g\)

⇔ \(x=0,1\left(mol\right)\)

➞ m_Fe= \(n.M=0,1.56=5,6\left(g\right)\)

m_Cu=\(n.M=0,1.64=6,4\left(g\right)\)

Vậy m_Cu=6,4 g

m_Fe=5,6 g