Question

hoà tan 13,65g hh X gồm Mg,Zn,Fe vào 160g dd H2SO4 24,5% thu đc 7,84l H2(đktc) và m g muối . tính m

Answer 1

\(n_{H_2}=\dfrac{7.84}{22.4}=0.35\left(mol\right)\)

\(n_{H_2SO_4}=\dfrac{160\cdot24.5\%}{98}=0.4\left(mol\right)\)

Ta có : 

\(n_{H_2SO_4}>n_{H_2}\Rightarrow H_2SO_4dư\)

\(n_{H_2SO_4\left(pư\right)}=n_{H_2}=0.35\left(mol\right)\)

BTKL:

\(m_{Muối}=m_{hh}+m_{H_2SO_4\left(bđ\right)}-\left(m_{H_2}+m_{H_2SO_4\left(dư\right)}\right)\)

\(=13.65+0.4\cdot98-0.35\cdot2-\left(0.4-0.35\right)\cdot98=47.25\left(g\right)\)