2Al(s)+6Hcl(aq)-----> 2AlCl3(s) + 3H2(g) Molar mass Al= 26.98 g/mol Molar mass HCl= 36.45g/mol Molar mass AlCl3=133.33 g/mol Molar mass H2= 2.00g/mol If 15.0 grams of aluminum completely react according to the provided equation and 1.54 grams of hydrogen are collected, what is the percent yield? Write in the CORRECT NUMBER OF SIGNIFICANT FIGURES. -------%
Added by Robert P.
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Given mass of Al = 15.0 grams Molar mass of Al = 26.98 g/mol \[ \text{Moles of Al} = \frac{\text{mass of Al}}{\text{molar mass of Al}} = \frac{15.0 \, \text{g}}{26.98 \, \text{g/mol}} = 0.556 \, \text{mol} \] Show more…
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