Here, the reaction is as follows
"H_2+O_2\\longrightarrow2H_2O"
Here given that mass of H2=8.0 g and mass of H2O= 9.0 g
moles of "H_2=\\frac{given mass}{molar mass}= \\frac{8}{2}= 4" mole and
moles of "O_2= \\frac{given mass}{molar masS}=\\frac{9}{32}" =0.2812
so here O2 is limiting reagent and
massof "H_2O = 18\\times.2812=5.06 g"
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