\(n_{HCl}=\frac{200.7,3}{100.36,5}=0,4\left(mol\right)\)
mHCl=\(\frac{200\cdot7,3\%}{100\%}\)=14,6(g)
nHCl=\(\frac{14,6}{36,5}\)=0,4(mol)
m HCl = 14,6(g)
=>n HCl= 14,6/36,5= 0,4 (mol)
mHCl = \(\frac{200.7,3\%}{100\%}\) = 14,6 g
nHCl = 14,6/36,5 = 0,4 mol