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Question
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Answer: C
π‘ Solution & Explanation
ο¨ο© ο¨ο© ο¨ο© ο¨ ο© ο¨ ο© 2 2 3 N g 3H g 2NH g Initial 1 3 0 At equil. 1 x 3 1 x 2x ο« ο ο οοοο οοοο ο¨ ο© 3 2 4 4 4 2 2NH H SO NH SO ο« οοοο οοοο Moles of 2 4 H SO required MV 250 1
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