calculate how many grams of heavy water are required to produce 345.0 mg of ND3(g).?

Question

Lithium nitride reacts with water to produce ammonia and lithium hydroxide according to the equation, Li3N(s)+3H2O(l) NH3(g)+3LiOH(aq)Heavy water is water with the isotope deuterium in place of ordinary hydrogen, and its formula is D2O. The above reaction can be used to produce heavy ammonia, ND3(g) according to the equation,Li3N(s)+3D2O(l) ND3(g)+3LiOD(aq)calculate how many grams of heavy water are required to produce 345.0 mg of ND3(g).The mass of deuterium, D, is 2.014 g/mol.

Mr. Toni Zulauf Sr. · Accepted Answer

(mass ND3, mg)(1g / 10mg)(mol/g ND3)(mole ratio D2O/ND3)(g/mol D2O) = mass D2O (345.0mg ND3)(1g / 10mg)(1mol ND3 / 20.0487g ND3)(3mol D2O / 1mol ND3)(20.0414g D2O /mol D2O)= 103.5g D2O

Mr. Toni Zulauf Sr. · Answer

(320.0mg ND3)(1g / 1000mg)(1 mol ND3 / 20.049g ND3)(3 mol D2O / 1 mol ND3)(20.027g / mol D2O) = .9589g D2OI got this answer correct on my online Sapling Learning homework for General Chemistry.

Mr. Toni Zulauf Sr. · Answer

ian, your answer is incorrect because the molar mass of g of D2O is incorrect, the correct molar mass is 20.027. if you recalculate it at the end with the correct molar mass the correct final answer should be 1.034g D2O. katie, you are given a different value, rather than 345.0mg ND3 you are given 320.0 so your answer will be different

Sigrid O'Kon

calculate how many grams of heavy water are required to produce 345.0 mg of ND3(g).?

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