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Amount of calcium oxide required, when it reacts with 852 g of P₄O₁₀ is
Options
(a) 100g
(b) 1008g
(c) 108g
(d) 1050g
Correct Answer:
1008g
Explanation:
6CaO + P₄O₁₀ →2Ca₃(PO₄)₂
6(40+ 16) 4x 31+ 10x 16
=(336) =284
284 g P₄O₁₀ required CaO=336g
852 g P₄O₁₀required CaO=(336x 852)/284=1008 g.
Related Questions: - Which of the following is used to prepare Cl₂ gas at room temperature from
- At S.T.P the density of CCl₄ vapours in g/L will be nearest to
- According to IUPAC nomenclature sodium nitroprusside is named as
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Question Type: Memory
(964)
Difficulty Level: Easy
(1008)
Topics: Basic Concepts of Chemistry
(94)
Subject: Chemistry
(2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Which of the following is used to prepare Cl₂ gas at room temperature from
- At S.T.P the density of CCl₄ vapours in g/L will be nearest to
- According to IUPAC nomenclature sodium nitroprusside is named as
- Which of the following shows linear structure
- The oxidation state of halogen atom is ‘+3’ in
Question Type: Memory (964)
Difficulty Level: Easy (1008)
Topics: Basic Concepts of Chemistry (94)
Subject: Chemistry (2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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