Phản ứng xảy ra:
\(hh_{kim.loai}+H_2SO_4\rightarrow muoi+H_2\)
Ta có:
\(n_{H2}=\frac{7,84}{22,4}=0,35\left(mol\right)=n_{H2SO4}\) (bảo toàn hidro)
BTKL; \(m_{hh\left(kim.loai\right)}+m_{H2SO4}=m_{muoi}+m_{H2}\)
\(\Leftrightarrow m+0,35.98=42,7+0,35.2\)
\(\Rightarrow m=8,9\left(g\right)\)