Answer:
223 g O₂
Explanation:
To find the mass of oxygen gas needed, you need to (1) convert moles Al to moles O₂ (via the mole-to-mole ratio from reaction coefficients) and then (2) convert moles O₂ to grams O₂ (via the molar mass). When writing your ratios/conversions, the desired unit should be in the numerator in order to allow for the cancellation of the previous unit. The final answer should have 3 sig figs because the given value (9.30 moles) has 3 sig figs.
4 Al + 3 O₂ ----> 2 Al₂O₃
^ ^
Molar Mass (O₂): 32.0 g/mol
9.3 moles Al 3 moles O₂ 32.0 g
------------------- x --------------------- x -------------------- = 223 g O₂
4 moles Al 1 mole
Answer:
Yes Concurred,but where is the question
Answer:
volume = 972.23ml
Explanation:
using general gas law
P1V1/T1 = P2V2/T2
765 x 585/293 = 443 x V2/282
1527.39 =443 x V2/282
1527.38 x 282 = 443 x V2
430695.78 = 443 x V2
V2 = 430695.68/443
V2 = 972.23mL
