Calcium carbide reacts with nitrogen and forms an important fertilizer. How many grams of this fertilizer can be formed when 6.4 g of calcium carbide is completely used for this process?

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Solution

$C a C_{2} + N_{2} \to C a C N_{2} + C$
Molecular weight of $C a C_{2}$ = $40 + (12 \times 2)$
=64 g
Molecular weight of $C a C N_{2}$ = $40 + 12 + (14 \times 2)$
=80 g
From the reaction 1 mole of $C a C_{2}$ forms 1 mole of $C a C N_{2}$ .
6.4 g of $C a C_{2}$ = $\frac{6.4}{64} = 0.1 m o l e$
So it will form 0.1 mole of $C a C N_{2}$ .
Amount of calcium cyanamide produced = $0.1 \times 80 = 8 g$

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Calcium carbide reacts with nitrogen and forms an important fertilizer. How many grams of this fertilizer can be formed when 6.4 g of calcium carbide is completely used for this process?

CaC2+N2→ CaCN2+ C Molecular weight of CaC2 = 40+(12×2) =64 g Molecular weight of CaCN2 = 40+12+(14×2) =80 g From the reaction 1 mole of CaC2 forms 1 mole of CaCN2. 6.4 g of CaC2 =6.464=0.1 mole So it will form 0.1 mole of CaCN2. Amount of calcium cyanamide produced =0.1×80=8 g