How many grams of ammonia must you start with to make 900.00 l of a 0.140 m aqueous solution of nitric acid? assume all the reac tions give 100% yield?
1 answer:
You need the set of reactions that goes from ammonia to nitric acid. <span>
1) 4NH3(g)+5O2(g)-->4NO(g)+6H2O(g)
2) 2NO(g)+O2(g)-->2NO2(g) 3) 3NO2(g)+H2O(l)-->2HNO3(aq)+NO(g) State the ratio of moles of HNO3 to NH3: 4 moles of NH3 produce 4 mole of NO, 4 moles of NO produce 4 moles of NO2 4 moles of NO2 produce 4 * (2 / 3) moles of HNO3 = 8/3 moles of HNO3. => (8/3) moles HNO3 : 4 moles NH3 Calculate the number of moles of HNO3 in 900.00 l of 0.140 M solution M = n / V => n = M * V = 0.140 M * 900.00 liter = 126 moles HNO3 Use proportions: (</span><span>8/3) moles HNO3 / 4 moles NH3 = 126 moles HNO3 / x => x = 126 moles HNO3 * 4 moles NH3 / (8/3 moles HNO3) = 189 moles NH3 Convert moles to grams: molar mass NH3 = 14 g/mol + 3 * 1g/mol = 17 g/mol mass in grams = number of moles * molar mass = 189 moles * 17 g/mol = 3213 g Answer: 3213 g. </span>
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