How many grams of Mo2S3 are needed to react completely with 10 liters of O2 gas at 27 °C and 0.750 atm.?
in the following reaction (molar mass Mo2S3 288.1 g/mole)
how would i solve this?
Update:(molar mass Mo2S3 288.1 g/mole)
in the following reaction:
2MO2S3 + 9O2 --> 6SO2 + 2MO2O3
how would i solve this?
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Ok well you would use PV=nRT, where
P=pressure (.750 atm),
V=volume (10 L),
n=moles (what you are looking for),
R=gas constant (.0821), and
T=temperature (27C-convert to Kelvin [+273]).
You need to find n by pluggin in your numbers into the equation, so it would look like this:
(0.750)(10)=n(0.0821)(300K)
Once you find n, you would multiply that by the g/moles (288.1) and there is your answer to the problem above. :)
*************DO NOT FORGET ABOUT SIG-FIGS********************
First you should stability the eqn: CH4 + 2O2 ---> CO2 + 2H2O Then bypass from g of CH4 --> mol CH4 --> mol O2 --> g O2 (32.08g CH4)(1mol CH4 / sixteen.04g CH4)(2mol O2 / 1mol CH4)(32.00g O2 / a million mol O2) = 128.0g O2 I Lu V Re Se Ar C H!