40g Na OH react with 60g H2S O 4? what reactant is limiting .0?
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You need to begin with the balanced equation such that:
`2NaOH + H_2SO_4 -> Na_2 SO_4 + 2H_2O`
You need first to evaluate the original ratio of moles of reactants `NaOH` and `H_2SO_4` , such that:
`40 g NaOH * (1mol NaOH)/(40 g NaOH) = 1 mol NaOH`
`60 gH_2SO_4 * (1mol H_2SO_4)/(98.1 g NaOH) = 0.61 mol H_2SO_4`
The mole ratio available is `(1 mol NaOH)/(0.61 mol H_2SO_4).`
You need to compare the mole ratio `(2 mol NaOH)/(1 mol H_2SO_4) =(2 mol NaOH)/(2*1/2 mol H_2SO_4)` to the available ratio `(1 mol NaOH)/(0.61 mol H_2SO_4)` .
Hence, evaluating the limiting reactant yields that 1 mol of `NaOH` was reacted by the `0.5` mol `H_2SO_4` , hence NaOH is limiting reactant.
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