When 40 grams of ammonium nitrate explode 14 grams of nitrogen and 8 grams of oxygen are formed. How many grams of water are formed?

Md Ariful Islam

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Md Ariful Islam

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You can calculate this by such reaction: NH₄NO₃ --gt; N₂ + frac12;O₂ + 2H₂O ==gt; 2NH₄NO₃ --gt; 2N₂ + O₂ + 4H₂O The law of conservation of mass solves this problem. If you start off with 40 grams you have to end up with 40 grams in a chemical reaction. If you produce 14 grams of N2 and 8 grams of O2, then add them together to get 22 grams. Hence m(N2) = 14 g; m(O2) = 8 g; In the sum it will be m(N2 + O2) = 14 + 8 = 22 grams Subtract the 22 from 40 grams and that gives you 18 grams left which is the water. In general it will be: m(H2O) = m(NH₄NO₃) - m(N2 + O2) = 40 - 22 = 18 grams Answer: 18 grams of water will be formed
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What mass of ammonium nitrate is required to lower the temperature of 400 mL of water from 25 oC to 1oC? The Hsoln for ammonium nitrate is 34 kJ/mol.

94.5 g

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The compound ammonium nitrate is a strong electrolyte. Write the reaction when solid ammonium nitrate is put into water. Include states of matter in your answer.

NH4(aq)+H2O(l)⇌NH3(aq)+H3O

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How many grams of ammonium nitrate need to be added to 500g of water so that the freezing point becomes -8.3degrees celcius?

Answer T=km T=8,3 k=1,84 (constant for water) m – molality 8,3/1,84=m=m(NH4NO3)/(M(NH4NO3)*0,5) m(NH4NO3)=8,3*80*0,5/1,84=180 g

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What mass of ammonium nitrate is required to lower the temperature of 400 mL of water from 25 oC to 1oC? The Hsoln for ammonium nitrate is 34 kJ/mol.

400 ml = 400 g of water ΔT = 25 – 1 = 24 o C C = 4.18 J/g∙C Q = cmΔT = 4.18 J/g∙C x 400 g x...

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If Ammonium Nitrate (NH4NO3) is 35% Nitrogen, what is the mass of Nitrogen in 2.5 kg of NH4NO3?

As Nitrogen comprises 35% of the mass of ammonium nitrate, 2.5 kg of NH4NO3 contains:m(N) = 2.5 kg × 35 % / 100% = 0.875 kg = 875 gAnswer: 875...

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How many gram does nitrogen (N) present in 100 Kg ammonium nitrate ((NH4NO3) fertilizer?

M (NH4NO3) = 80.04 g/moln (NH4NO3) = m / M = 100 000 / 80.04 = 1249.33 molThe ratio of moles of N to moles of NH4NO3 is 2:1. Therefore,...

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If you start with 25.0 grams of lead (II) nitrate and 15.0 grams of sodium iodide, how many grams of sodium nitrate can be formed?

Pb(NO3)2 + 2NaI = 2NaNO3 + PbI2n = m / MM (Pb(NO3)2) = 331.2 g/molM (NaI) = 150 g/molM (NaNO3) = 85 g/moln (Pb(NO3)2) = 25 / 331.2 = 0.08...

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According to the following reaction, how many grams of ammonium chloride are needed to form 30.8 grams of hydrochloric acid? ammonium chloride (aq) ammonia (g) + hydrochloric acid (aq)

. m (HCl) =30.8 g .m (NH4Cl)=? The reaction is NH4Cl= NH3 + HCl Let's find the amount of substance (HCl)= m(HCl)/ Mr(HCl) Mr (HCl)= Ar(H)+ Ar(Cl)=1+35.5= 36.5 . n=30.8/36.5=0.84...

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A 0.4230 g sample of impure sodium nitrate was heated, converting all the sodium nitrate to .1235 g of sodium nitrate and oxygen gas. Determine the percent of sodium nitrate in the original sample.

Answer:2HNO3 = 2HNO2 + O2m= 85 x 0.1235 / 69= 0.1521W%= 0.1521 / 0.4230 = 36%

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Ammonium nitrate + sodium hydroxide yields ammonia + water + sodium nitrate

NH4NO3 + NaOH ---> NH3 + NaNO3 + H2O NH4+(aq) + OH-(aq) --> NH3(g )+ H2O(l)

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