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Amir Qadim Bajwa
Asked: 2022-10-29T02:41:14+05:30 In:

Limiting reactant: What mass of water `(H_(2) O)` can be formed by the reaction of `3.00g` of `H_(2) (g)` with `29.0g` of `O_(2)(g)`?
Strategy: Using the balanced equation
`{:(2H_(2)(g),+,O_(2)(g),,rarr2H_(2)O(l)),(2mol,,1mol,,2mol),(2(2.00g),,32.0g,,2(18.0g)):}`
find out the numbers fo moles of each reactatn required to react with the other. Using the given masses, calculate the number of moles of each reactant. Finally, identify the limiting recatant and base the rest of the calculate on it.

Limiting reactant: What mass of water `(H_(2) O)` can be formed by the reaction of `3.00g` of `H_(2) (g)` with `29.0g` of `O_(2)(g)`?
Strategy: Using the balanced equation
`{:(2H_(2)(g),+,O_(2)(g),,rarr2H_(2)O(l)),(2mol,,1mol,,2mol),(2(2.00g),,32.0g,,2(18.0g)):}`
find out the numbers fo moles of each reactatn required to react with the other. Using the given masses, calculate the number of moles of each reactant. Finally, identify the limiting recatant and base the rest of the calculate on it.
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  1. Step 1: Calculate the moles of each reactant using the formula,
    `n = (“Mass”)/(“Molar mass”)`
    `? “mol” H_(2) = 3.00g H_(2) xx (1 “mol” H_(2))/(2.00g H_(2)) = 1.50 mol H_(2)`
    `? “mol” O_(2) = 29.0g O_(2) xx (1 “mol” O_(2))/(32.0g O_(2)) = 0.906 “mol” O_(2)`
    Step 2: Using the balanced equation, determine the required moles of each reactant.
    The balalnced equation tells us that for every 1 mol `O_(2)` to react completely, 2 mol `H_(2)` is needed. Similarly for every 2 mol `H_(2)` to react completely, 1 mol `O_(2)` is needed, i.e.,
    `(n_(H_(2)))_(“required”) = 2(n_(O_(2)))_(“given”)`
    `(n_(O_(2)))_(“required”) = (1)/(2) (n_(H_(2)))_(“given”)`
    Step 3: Finding the limiting reacant.
    We see that `2xx(0.906)` or `1.81` mol of `H_(2)` are required to consume `O_(2)` completely but we have just 1.50 mol `H_(2)` , so `H_(2)` is the limiting rectant. Alternatively, we observe that only `(1)/(2) (1.50)` or `0.75` mol `O_(2)` is required to consume `H_(2)` completely but we have 0.906 mol `O_(2)` , so we see again that `H_(2)` is the limiting reacant while `O_(2)` is the excess reacant.
    Step 4: The reactiion must stop when the limiting reactant, `H_(2)` is used up , so we base the calculation on `H_(2)`.
    Either we use unitary method or proceed through the mole concept :
    Unitary method : Accroding to the balanced equation, `4g` of `H_(2)` yields `36g` of `H_(2)O` , thus, `1 g` of `H_(2)` will yield `36//4 g` of `H_(2) O` and `3g` fo `H_(2)` will yield `3xx36//4g = 27g H_(2) O`
    Mole concept :
    `underset(H_(2))(“g of”)rarr underset(H_(2))(“mol of”)rarr underset(H_(2)O)(“mol of”)rarr underset(H_(2)O)(“g of”)`
    `? g H_(2) O = 3.00g H_(2) = (1 “mol” H_(2))/(2.00g H_(2)) xx (2 “mol” H_(2) O)/(2 “mol” H_(2))`
    `xx (18.0 g H_(2) O)/(1 “mol” H_(2) O)`
    `= 27.0g H_(2) O`

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