\(n_{\left(NH_4\right)_2SO_4}=\dfrac{33}{132}=0,25\left(mol\right)\)
Ta có: \(n_H=8n_{\left(NH_4\right)_2SO_4}=8\times0,25=2\left(mol\right)\)
n(NH4)2SO4=\(\dfrac{m_{\left(NH4H\right)2SO4}}{M_{\left(NH4\right)2SO4}}\)=\(\dfrac{33}{132}\)=0,25mol
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