Answers
Best Answer: AlBr3 + 3 NaOH → Al(OH)3 + 3 NaBr
a.
Supposing you meant "How many moles of Sodium bromide can be formed..."
(1.55 mol AlBr3) x (3/1) = 4.65 mol NaBr
b.
(4.65 mol NaOH) x (1/3) = 1.55 mol Al(OH)3
CH4 + 2 O2 → CO2 + 2 H2O
a. (3.5 x 10^-4 mol CO2) x (1/1) = 3.5 x 10^-4 mol CH4
b. (3.5 x 10^-4 mol CO2) x (2/1) = 7.0 x 10^-4 mol O2
c. (7.0 x 10^-4 mol O2) x (2 atoms / molecule) = 14 x 10^-4 mol O
I hope I helped
