The ratio of moles of AI(NO3)3 to moles Na2CO3 in the reaction is 2:3
In the reaction 2 moles of AI(NO3)3 reacted with 3 moles Na2CO3 so the ratio of the moles is 2 is to 3 represented as 2:3
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if 105 grams burns completely
therefore
105 ×22.4/48=49
PV = nRT
1.026atm * 42l = n * 0.0821L atm/mol K * 305K
n=1.72.
1.72 molecules.