CH4 +2 O2 = CO2 + 2 H2O
You burned 15.0 grams of methane (CH4) in an excess of oxygen and formed 23.8 grams of water. What was your percent yield, rounded to 3 sig-figs?
Molar mass of methane "=16g\/mol"
Molar mass of water "=18g\/mol"
Moles of methane burned"=\\frac{15}{16}=0.9375" moles
Moles of water "=0.9375\u00d72"
"=1.875"
Mass of water "=1.875\u00d718"
"=33.75g"
"\\%" yield "=\\frac{23.8}{33.75}\u00d7100\\%"
"=70.5\\%"
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