Rust is a reddish brown solid with the chemical formula Fe2O3 · nH2O. To determine the number of molecules of water of crystallization in a formula unit of a rust sample, 40.0 cm3 of 1.5M hydrochloric acid was added to 1.89 g of.... (Details below)?

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Answer 1 · 2017-08-25T14:31:29

#sf(0.0424)#

#sf(n=5.97)#

The iron(III) oxide in the rust reacts with the acid which is INXS:

#sf(Fe_2O_(3(s))+6HCl_((aq))rarr2FeCl_(3(aq))+3H_2O_((l)))#

The initial no. of moles of HCl added is given by:

#sf(nHCl_("init")=cxxv=1.5xx40.0/1000=60.0xx10^(-3))#

The unreacted HCl is titrated against 1.20 M NaOH:

#sf(HCl_((aq))+NaOH_((aq))rarrNaCl_((aq))+H_2O_((l)))#

#sf(nOH^(-)=cxxv=1.20xx14.65/1000=17.58xx10^(-3))#

From the equation we can say that:

#sf(nHCl_("rem")=17.58xx10^(-3))#

#:.# the no. of moles of HCl consumed is given by:

#sf(nHCl=(60.0-17.58)xx10^(-3)=42.42xx10^(-3))#

This is the answer to the 1st part of the question.

From the equation we can say that:

#sf(nFe_2O_3=(42.42xx10^(-3))/(6)=7.07xx10^(-3))#

Mass #sf(Fe_2O_3=nxxM_r=7.07xx10^(-3)xx159.688=1.12899color(white)(x)g)#

#:.#Mass #sf(H_2O=1.89-1.12899=0.76101color(white)(x)g)#

The no. moles of #sf(H_2O)# is given by:

#sf(nH_2O=m/M_r=0.76101/18.0153=0.04224)#

Ratio by moles #sf(Fe_2O_3:H_2OrArr)#

#sf(0.00707:0.04224)#

=#sf(1:5.97)#

So we can write the formula:

#sf(Fe_2O_3.(H_2O)_(5.97))#

I'll leave it unrounded as rust is not a stoichiometric compound with regard to the water present.