Gọi số mol Ca(NO3)2, CuSO4 là a, b (mol)
=> \(\left\{{}\begin{matrix}164a+160b=145,2\left(g\right)\\a+b=\dfrac{5,4.10^{23}}{6.10^{23}}=0,9\left(mol\right)\end{matrix}\right.\)
=> a = 0,3 (mol); b = 0,6 (mol)
\(\left\{{}\begin{matrix}m_{Ca\left(NO_3\right)_2}=0,3.164=49,2\left(g\right)\\m_{CuSO_4}=0,6.160=96\left(g\right)\end{matrix}\right.\)
\(n_O=0,3.6+0,6.4=4,2\left(mol\right)\Rightarrow m_O=4,2.16=67,2\left(g\right)\)
\(n_{Ca}=0,3\left(mol\right)\Rightarrow m_{Ca}=0,3.40=12\left(g\right)\)
\(n_N=0,3.2=0,6\left(mol\right)\Rightarrow m_N=0,6.14=8,4\left(g\right)\)
\(n_{Cu}=0,6\left(mol\right)\Rightarrow m_{Cu}=0,6.64=38,4\left(g\right)\)
\(n_S=0,6\left(mol\right)\Rightarrow m_S=0,6.32=19,2\left(g\right)\)
