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дано

m(Fe(OH)3 - 2.14 g

W(HCL) = 20%

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m(ppa HCL) -?

2Fe(OH)3+6HCL-->2FeCL3+6H2O

M(Fe(OH)3) = 107 g/mol

n(Fe(OH)3) = m/M = 2.14 / 107 = 0.02 mol

2n(Fe(OH)3) = 6n(HCL)

n(HCL) = 6*0.02 / 2 = 0.06 mol

M(HCL) = 36.5 g/mol

m(HCL) = n*M = 0.06 * 36.5 = 2.19  g

m( ppa HCL) = 2.19 * 100% / 20% = 10.95 g

ответ 10.95 г