C4h10+6.5o2=4co2+5h2o
moles of butane=1.92/58=0.0331 moles
moles of water=0.1655 moles\
as the butane and water has 1 is to 5 molar ratio
0.1655=mass/18
mass=2.98 g
mass of water produced = 2.98 g
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Answer:
There are 0,011 moles of hydrogen gas.
Explanation:
We use the ideal gas formula, with the constant R = 0.082 l atm / K mol. The STP conditions are : 1 atm pressure and 273 K temperature. Solve for the formula, n (number of moles):
PV=nRT ---> n= (PV)/(RT)
n= (1 atm x 0,25 L)/ (0,082 l atm/ K mol x 273 K)
<em>n= 0,011 mol</em>