Asked by Victor
You want to make 40 mL of gas from some aluminum carbonate. The temperature of the lab is 23 C and the rpessure is 760 torr. If you collect the gas over water, what mass of reactant is necessary?
Answers
Answered by
bobpursley
Al2(CO3)3+3H2O⟶2Al(OH)3+3CO2
Ok, for each mole of aluminum carbonate, one gets 3 moles of CO2.
Using the ideal gas law
PV=nRT
n=PV/RT=760*.040/(296*62.364)
n=.00164682674 check that.
so moles aluminum carbonate=n/3
then convert those moles to grams
grams=moles*formulmassaluminiumcarbonate.
Ok, for each mole of aluminum carbonate, one gets 3 moles of CO2.
Using the ideal gas law
PV=nRT
n=PV/RT=760*.040/(296*62.364)
n=.00164682674 check that.
so moles aluminum carbonate=n/3
then convert those moles to grams
grams=moles*formulmassaluminiumcarbonate.
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