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Mathematics 15 Online
OpenStudy (anonymous):

What mass of oxygen is needed for the complete combustion of 7.70×10−3gg of methane

OpenStudy (anonymous):

\[CH_4+2O_2~~\implies~~CO_2+2H_2O\] which means for every mole of methane, you need 2 moles of oxygen for complete combustion. Molar masses: \[\begin{array}{c|c} CH_4& 12+4(1)=16\frac{\text{g}}{\text{mol}}\\ \hline O_2& 2(16)=32\frac{\text{g}}{\text{mol}} \end{array}\] \[(7.70\times10^{-3}\text{g})\times\frac{1\text{ mol}}{16\text{ g}}=x\text{ moles of }CH_4\] which means you'll need \(2x\) moles of \(O_2\). Convert to mass.

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