A crop of wheat uses up to 20 kg of nitrogen per hectare of soil. What is the mass of the fertilizer calcium nitrate required to replace the nitrogen in a 10-hectare field?
$(C a = 40; N = 14, O = 16)$

1000 kg

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1171 kg

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1271 kg

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1371 kg

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Solution

The correct option is C 1171 kg
Molecular mass of $C a (N O_{3})_{2} = 40 + 2 (14) + 6 (16)$
$= 164 g$
1 hectare of roll removes 20 Kg of Nitrogen.
10 hectare of roll removed $20 \times 10 = 200 K g N_{2}$
$= 200 \times 1000 g$ of nitrogen.
28g of $N_{2}$ is replaced by 164g of $C a (N O_{3})_{2}$
[as calcium nitrate has 28g of $H_{2}$ ]
$200 \times 1000 g$ of $N_{2}$ is replaced by $\frac{164}{28} \times 200 \times 1000$
$117.14 \times 10^{3} g$
$= 11747 K g$ of $C a (N O_{3})_{2}$
Mass of fertilizer required $= 1171 K g$

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A crop of wheat uses up to 20 kg of nitrogen per hectare of soil. What is the mass of the fertilizer calcium nitrate required to replace the nitrogen in a 10-hectare field? (Ca=40; N=14, O=16)

A crop of wheat uses up to 20 kg of nitrogen per hectare of soil. What is the mass of the fertilizer calcium nitrate required to replace the nitrogen in a 10-hectare field?
$(C a = 40; N = 14, O = 16)$