The balanced equation for the decomposition of ammonium nitrate is
2 NH₄ NO₃(s) → 2 N₂(g)+O₂(g)+4 H₂ O(g)
(a) Select the mole ratios that relate moles of each product to the reactant NH₄ NO₃.
1 mol O₂/2 mol NH₄ NO₃ 4 mol H₂ O/2 mol NH₄ NO₃
2 mol NH₄ NO₃/4 mol H₂ O 2 mol N₂/2 mol NH₄ NO₃
2 mol NH4 NO₃/1 mol O₂ 2 mol NH₄ NO₃/2 mol N₂
(b) How many moles of each product would form if 1.00 mol of NH₄ NO₃ reacts?
Moles of N₂=_______ mol N₂
Moles of O₂=_______ mol O₂
Moles of H₂ O=_______ mol H₂ O
(c) How many moles of each product would form if 18.2 mol of NH₄ NO₃ react?
Moles of N₂=_______ mol N₂
Moles of O₂=_______ mol O₂
Moles of H₂ O=_______ molH₂ O
The balanced equation for the decomposition of ammonium nitrate is 2NH4NO3(s) → 2N2(g) + O2(g) + 4H2O(g) (a) Select the...
Ammonium nitrate decomposes explosively upon heating according to the following balanced equation: 2NH4NO3(s)→2N2(g)+O2(g)+4H2O(g) Calculate the total volume of gas (at 122 ∘C and 730 mmHg ) produced by the complete decomposition of 1.37 kg of ammonium nitrate.
Ammonium nitrate decomposes explosively upon heating according to the following balanced equation: 2NH4NO3(s)→2N2(g)+O2(g)+4H2O(g) Calculate the total volume of gas (at 119 ∘C and 761 mmHg ) produced by the complete decomposition of 1.71 kg of ammonium nitrate.
12) Ammonium nitrate decomposes explosively upon heating according to the following balanced equation: 2NH4NO3(s)→2N2(g)+O2(g)+4H2O(g) Calculate the total volume of gas (at 127 ∘C and 733 mmHg ) produced by the complete decomposition of 1.54 kg of ammonium nitrate.
The balanced equation for the decomposition of TNT, C₇ H₅(NO₂)₃, is2 C₇ H₅(NO₂)₃(s) → 7 C(s)+7 CO(g)+3 N₂(g)+5 H₂ O(l)How many moles of each product would form if 8.32 mol of C₇ H₅(NO₂)₃ react?_______ mol C_______ mol CO_______ mol N₂_______ mol H₂ O
Ammonium nitrate can decompose explosively when heated according to the following equation: 2NH4NO3(s) → 2N2(g)+4H2O(g)+O2(g) How many liters of gas would be formed at 430 Celcius and 1.00 atm pressure from an explosion of 430 g of NH4NO3?
Lead(II) nitrate and ammonium iodide react to form lead(II) iodide and ammonium nitrate according to the reaction Pb(NO3), (aq) + 2NH I(aq) PbI,() +2 NH NO, (aq) What volume of a 0.150 M NH I solution is required to react with 409 mL of a 0.100 M Pb(NO3), solution? volume: How many moles of Pbly are formed from this reaction? moles: mol Pbl,
Lead(II) nitrate and ammonium iodide react to form lead(II) iodide and ammonium nitrate according to the reaction Pb(NO3), (aq) + 2NH I(aq) → Pb1,(s) + 2 NH, NO, (aq) What volume of a 0.310 M NH I solution is required to react with 193 mL of a 0.680 M Pb(NO3)2 solution? volume: How many moles of Pbly are formed from this reaction? moles: mol PbI,
Lead(II) nitrate and ammonium iodide react to form lead(II) iodide and ammonium nitrate according to the reaction Pb(NO3), (aq) + 2 NH,I(aq) + PbI,(s) + 2 NH, NO3(aq) What volume of a 0.150 M NH4I solution is required to react with 591 mL of a 0.540 M Pb(NO3)2 solution? volume: mL How many moles of Pbly are formed from this reaction? moles: mol PbI2
Lead(II) nitrate and ammonium iodide react to form lead(II) iodide and ammonium nitrate according to the reaction Pb(NO3)2(aq) + 2NH,I(aq) — Pl(s) + 2NH, NO, (aq) What volume of a 0.270 M NH I solution is required to react with 487 mL of a 0.680 M Pb(NO3)2 solution? volume: How many moles of Pbly are formed from this reaction? moles: mol Pbl2
Lead(II) nitrate and ammonium iodide react to form lead(II) iodide and ammonium nitrate according to the reaction Pb(NO )2 (aq) + 2NH, l(aq) — Pbl (8) + 2NH, NO, (aq) What volume of a 0.690 M NH I solution is required to react with 883 mL of a 0.120 M Pb(NO), solution? volume: 2538.63 How many moles of Pbly are formed from this reaction? moles: 0.60927 mol Pbl: