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You must use 134 g O₂ to produce 118 g H₂O.
M_r: 32.00 18.02
2H₂ + O₂ ⟶ 2H₂O
Moles of H₂O = 150.9 g H₂O × (1 mol H₂O/18.02 g H₂O) = 8.374 mol H₂O
Moles of O₂ = 8.374 mol H₂O × (1 mol O₂/2 mol H₂O) = 4.187 mol O₂
Mass of O₂ = 4.1877 mol O₂ × (32.00 g O₂/1 mol O₂) = 134 g O₂
Answer : 13.64 mL
Explanation : The reaction requires 13.64 mL of NaOH
From the reaction we know that oxalic acid requires two moles of NaOH so we have to calculate the number of moles of both.
0.150 L X 0.1500 moles/L = 2.25 X
So the moles of NaOH will be 2 X (2.25 X ) = 4.5 X
Now, we know the concentration of NaOH as 0.330 M
4.5 X moles / 0.330 moles/L = 0.01364 L or 13.64 mL
So, the volume of NaOH needed to neutralize oxalic acid will be 13.64 mL