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mno2+4hcl=cl2+mncl2+2h2o if you were given 145 g of HCl how many grams of MnCl2 Could you Theoretically produce if you reacted with the excess MnO2

Sagot :

The first step is to use the molecular weight of HCl to convert 145g to moles of HCl (mw=36.458g/mol):

[tex]145g\cdot\frac{mol}{36.458g}=3.97mol[/tex]

Now, use the ratio of the coefficients of MnCl2 to HCl, to find how many moles of MnCl2 are produced with this amount of HCl:

[tex]3.97molHCl\cdot\frac{1molMnCl_2}{4molHCl}=0.9925molMnCl_2[/tex]

Use the molecular weight of MnCl2 to convert the amount of moles produced to grams:

[tex]0.9925molMnCl_2\cdot\frac{125.844g}{molMnCl_2}=124.9g[/tex]

124.9g of MnCl2 are produced.

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