Câu 55:
\(n_{NO_2}=\dfrac{6,72}{22,4}=0,3(mol)\\ a,Fe+6HNO_3\to Fe(NO_3)_3+3NO_2\uparrow+3H_2O\\ b,n_{Fe}=0,1(mol)\\ \Rightarrow m_{Fe}=0,1.56=5,6(g)\)
Câu 56:
\(n_{NO_2}=\dfrac{0,15}{22,4}=0,15(mol)\\ a,Al+6HNO_3\to Al(NO_3)_3+3NO_2\uparrow+3H_2O\\ b,n_{Al}=0,05(mol)\\ \Rightarrow m_{Al}=0,05.27=1,35(g)\)
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