Respuesta :
Answer:
Amount of methane needed = 4 g
Explanation:
Mass of water = 9
Molar mass of water = 18 g/mol
[tex]Mol = \frac{Mass\ in\ g}{Molar\ mass}[/tex]
No. of moles of water = [tex]\frac{9}{18} = 0.5\ mol[/tex]
Mass of carbon dioxide = 11
Molar mass of carbon dioxide = 44 g/mol
No. of moles of carbon dioxider = [tex]\frac{11}{44} = 0.25\ mol[/tex]
Mass of oxygen = 16
Molar mass of oxygen (O2) = 32 g/mol
No. of moles of carbon dioxider = [tex]\frac{16}{32} = 0.5\ mol[/tex]
[tex]CH_4 + 2O2 \rightarrow CO_2 + 2H_2O[/tex]
From reaction coefficient,
2 mol of oxygen reacts with 1 mol of methane to form 1 mol of carbon dioxide and 2 mol of water.
or, 1 mol of oxygen reacts with 0.5 mol of methane to form 0.5 mol of carbon dioxide and 1 mol of water.
SO, 0.5 mol of oxygen reaction with 0.25 mol of methane to form 0.25 mol of carbon dioxide and 0.5 mol of water.
No. of mol of methane needed = 0.25 mol
Molar mass of methane = 16
Mass of methane in g = No. of mol × Molar mass
= 0.25 × 16
= 4 g
When methane (CH4) is burned with oxygen, 4 g of methane were needed for the reaction.
The equation of the reaction is;
CH4(g) + 2O2(g) -------> CO2(g) + 2H2O(g)
Number of moles in 11g of CO2 = 11g/44g/mol = 0.25 moles of CO2
Number of moles in 16 g of O2 = 16 g/32g/mol = 0.5 moles
Since the reaction is 2:1
2 mole of O2 yields 1 mole of CO2
x moles of O2 yields 0.25 moles of CO2
x = 2 mole * 0.25 moles/1 mole
= 0.5 moles of O2
Also, the reaction between methane and oxygen is 1:2
1 mole of CH4 reacts with 2 moles of oxygen
x moles of methane reacts with 0.5 moles of O2
x = 1 mole * 0.5 moles/ 2 moles
x = 0.25 moles of CH4
Mass of 0.25 moles of CH4 = 0.25 moles * 16 g/mol = 4 g of CH4
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