Maybe this example could help you to understand this problem.
https://image.slidesharecdn.com/121howmanyatoms-091201144624-phpapp02/95/12-1-how-many-atoms-17-728....
Answer:
5.625 moles of oxygen, O₂.
Explanation:
The balanced equation for the reaction is given below:
4Al + 3O₂ —> 2Al₂O₃
From the balanced equation above,
4 moles of Al reacted with 3 moles of O₂.
Finally, we shall determine the number of mole of O₂ required to react with 7.5 moles of aluminum, Al. This can be obtained as illustrated below:
From the balanced equation above,
4 moles of Al reacted with 3 moles of O₂.
Therefore, 7.5 moles of Al will react with = (7.5 × 3)/4 = 5.625 moles of O₂.
Thus, 5.625 moles of O₂ is needed for the reaction.
Should be 1.8L.
2 moles of hydrogen react with 1 mole of oxygen. If 2 moles of hydrogen is 3.6L, 1 mole of oxygen should be 1.8L.
Answer: it will take 89.93secs
Explanation:Please see attachment for explanation
Answer:H2=11.4g
CH4=28.6g
Explanation:The complete combustion of the two gases can be represented by a balanced reaction below
1. CH4 +2O2___CO2+2H2O
2.2H2+O2___2H2O
Combining the two we have CH4 +2H2+3O2___
CO2+4H2O
Since the mixture contains 40gof CH4 and 2, therefore 20g of CH4 and 8g of H2 combines.
Calculated from their molecular Mass i.e CH4 12+4×2)=20 and 2H2= 2×2×2=8g
Mass of CH4=20/28×40=28.6g
2H2=8/28×40=11.4g
