Limiting reagent problems — Hinglish notes

1.3.3 · Chemistry › Chemical Reactions & Stoichiometry

Limiting reagent KIYA hota hai?

YEH KYON hota hai? (first principles)

Ek balanced equation jaise ki

$N_{2} + 3 H_{2} \to 2 N H_{3}$

ek fixed recipe hai: 1 mole $N_{2}$ ko exactly 3 moles $H_{2}$ chahiye. Agar tum $H_{2}$ aur $N_{2}$ kisi bhi aisi ratio mein supply karo jo $3 : 1$ se alag ho, toh unme se ek doosre se pehle "khatam" ho jaayega. Molecules whole-number ratios mein react karte hain (Dalton), isliye reaction ruk jaata hai jis instant ek species zero hit karti hai — baaki sab waise hi baith jaata hai.

KAISE solve karein — universal 4-step method

Steps:

Equation balance karo (coefficients $ν_{i}$ non-negotiable hain).
Diya hua har mass/volume moles mein convert karo: $n = m / M$ .
Har reactant ke liye $R_{i} = n_{i} / ν_{i}$ compute karo → sabse chhota limiting hai.
Product paane ke liye (aur bacha hua excess nikalne ke liye) limiting reagent se scale karo.

Worked Example 1 — classic ammonia synthesis

Q: $28 g$ $N_{2}$ , $10 g$ $H_{2}$ se react karke $N H_{3}$ banata hai. Limiting reagent, $N H_{3}$ ki mass, aur bacha hua excess nikalo.

$N_{2} + 3 H_{2} \to 2 N H_{3}$

Step 1 — moles. $n_{N_{2}} = \frac{28}{28} = 1 mol, n_{H_{2}} = \frac{10}{2} = 5 mol$ Yeh step kyon? Recipe particles count karti hai; pehle grams se moles mein jaana zaroori hai.

Step 2 — ratios $R_{i}$ . $R_{N_{2}} = \frac{1}{1} = 1, R_{H_{2}} = \frac{5}{3} = 1.67$ Yeh step kyon? Jo kam "reaction sets" fund kar sake woh pehle khatam hoga. $1 < 1.67$ , isliye == $N_{2}$ limiting hai==.

Step 3 — limiting reagent se product. $n_{N H_{3}} = n_{N_{2}} \times \frac{2}{1} = 1 \times 2 = 2 mol \Rightarrow m = 2 \times 17 = 34 g$ Yeh step kyon? Yield sirf limiting reagent decide karta hai; $N H_{3}$ ke saath uska mole-ratio use karo.

Step 4 — bacha hua $H_{2}$ . $H_{2} used = 1 \times 3 = 3 mol; leftover = 5 - 3 = 2 mol = 4 g$ Yeh step kyon? Excess reagent gayab nahi hota; jo consume hua use subtract karo.

Worked Example 2 — "badi dikhne waali" masses tumhe ullu banati hain

Q: $50.0 g$ Al, $80.0 g$ $O_{2}$ se react karta hai. Kaun limiting hai? ( $M_{A l} = 27$ , $M_{O_{2}} = 32$ )

$4 A l + 3 O_{2} \to 2 A l_{2} O_{3}$

Moles: $n_{A l} = 50/27 = 1.85 mol$ , $n_{O_{2}} = 80/32 = 2.50 mol$ .

Ratios: $R_{A l} = \frac{1.85}{4} = 0.463, R_{O_{2}} = \frac{2.50}{3} = 0.833$ Sabse chhota $R_{A l}$ hai ⇒ Al limiting hai chahe $O_{2}$ ki mass zyada ho. Yeh step kyon? Badi mass = badi supply nahi, jab coefficient aur molar mass se divide karo toh.

Product: $n_{A l_{2} O_{3}} = 1.85 \times \frac{2}{4} = 0.926 mol \Rightarrow m = 0.926 \times 102 = 94.4 g$ .

Worked Example 3 — percent yield ka connection

Q: Example 1 mein, sirf $28 g$ $N H_{3}$ actually collect hoti hai. Percent yield kya hai?

$% yield = \frac{28}{34} \times 100 = 82.4%$ Yeh step kyon? Theoretical yield ( $34 g$ ) limiting reagent se aayi thi; actual woh hai jo lab mein milta hai.

Common mistake Common galat ideas ko seriously lete hue unhe sahi karna

Galat idea 1: "Jis reactant ke paas kam grams hain woh limiting hai." Yeh sahi kyun lagta hai: kam cheez ⇒ jaldi khatam, intuitive lagta hai. Fix: Moles mein convert karo aur coefficient se divide karo. Example 2 mein Al ki mass kam thi aur woh limiting bhi tha — lekin woh coincidence tha; generally grams jhooth bolte hain.

Galat idea 2: "Jis reactant ke paas kam moles hain woh limiting hai." Yeh sahi kyun lagta hai: moles sahi unit hai, toh unhe compare karna kaafi lagta hai. Fix: Phir bhi stoichiometric coefficient se divide karna padega. Zyada moles waala reactant phir bhi limiting ho sakta hai agar uska coefficient bada ho (zyada ki zaroorat hai per reaction).

Galat idea 3: "Excess reagent se product compute karo." Yeh sahi kyun lagta hai: uske paas 'zyada' hai, toh lagta hai woh zyada product drive karta hai. Fix: Excess bacha hua hai; woh bina partner ke react nahi kar sakta. Hamesha limiting reagent se scale karo.

Recall Feynman: ek 12-saal ke bacche ko samjhao

Socho tum toy cars bana rahe ho. Har car ko 4 wheels aur 1 body chahiye. Tumhare paas 20 wheels hain lekin sirf 3 bodies. Tum sirf 3 cars bana sakte ho, kyunki 3 ke baad koi body nahi bachi — chahe abhi bhi 8 wheels pade hoon. Bodies ne tumhe "limit" kiya. Chemistry mein, jo reactant pehle khatam ho woh boss hai jo decide karta hai tum kitna bana sakte ho; baaki sab waise hi pada rehta hai, apne partner ke bina bekar.

Flashcards

Limiting reagent define karo

Woh reactant jo pehle completely consume ho jaata hai; yahi maximum product ko cap karta hai.

Grams nahi, moles kyun compare karte hain?

Reactions whole-number particle (mole) ratios mein hoti hain; grams molar mass ki wajah se alag hote hain aur galat raasta dikhate hain.

Kaun si quantity

R_{i}

limiting reagent identify karti hai?

R_{i} = n_{i} / ν_{i}

(moles ÷ stoichiometric coefficient); sabse chhota

R_{i}

limiting hai.

Theoretical yield kis reagent se compute karte hain?

Sirf limiting reagent se.

N_{2} + 3 H_{2} \to 2 N H_{3}

mein 1 mol

N_{2}

, 5 mol

H_{2}

ke saath, kaun limiting hai?

N_{2}

(

R = 1 < 5/3

Percent yield ka formula

(actual yield / theoretical yield) × 100%.

Kya excess reagent product ki amount affect karta hai?

Nahi; sirf limiting reagent product ki amount set karta hai. Excess bacha rehta hai.

Bacha hua excess reagent kaise nikaalte hain?

(initial moles) − (limiting reagent ke saath react karne mein consume hue moles), phir grams mein convert karo.

Connections

Balancing chemical equations — coefficients $ν_{i}$ yahan se aate hain.
Mole concept and Avogadro number — kyun moles currency hai.
Stoichiometric calculations — general mass↔mole↔mass conversions.
Percent yield and purity — direct downstream application.
Concentration and molarity — solution-phase limiting reagent problems.