標題:
GASES stoichiometry
發問:
CO (g) + 2H2(g)--->CH3OH(g)hydrogen at STP flows into a reactor at a rate of 16.0L/min. Carbon monoxide at STP flows into the reactor at a rate of 25.0 L/min. If 5.30gof methanoal is produced pre minute, what is the percent yield if the reaction?ANS:46.5%How to get it...please helpTHANKSSSS! 顯示更多 CO (g) + 2H2(g)--->CH3OH(g) hydrogen at STP flows into a reactor at a rate of 16.0L/min. Carbon monoxide at STP flows into the reactor at a rate of 25.0 L/min. If 5.30g of methanoal is produced pre minute, what is the percent yield if the reaction? ANS:46.5% How to get it... please help THANKSSSS!
最佳解答:
From the data that molar volume of any gas under STP = 22.4 dm3: No. of moles of hydrogen flowing into the reactor = 0.714 mole No. of moles of hydrogen flowing into the reactor = 1.116 moles So according to the equation, hydrogen is the limiting reactant and, the theoretical no. of moles of methanol produced per minute is 0.357 mole in accordance to the equation. So theoretical mass of methanol produced per minute = 0.357 x 32 = 11.43 g Hence percentage yield = (5.3/11.43) x 100% = 46.5%
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