Calculate the amount of heat needed to decompose 25.0g of ammonia NH3 into its elements?
Moles NH3 = 25.0 g / 17.0 g/mol = 1.47 mol NH3 46.19 kJ/mol NH3 X 1.47 mol NH3 = 67.9 kJ heat required
Calculate the amount of heat needed...?
Two steps: 1. How many moles of NH3 are in 25g of NH3? 25g NH3 x 1 mol / 17g = 1.47 mol NH3 (17 comes from the periodic table: N + 3H = 14 + 3(1) = 17 ) 2. How much energy is needed to decompose that many moles? 1 mole takes 46kJ, so 1.47 moles will take 1.47 times as much. 46.19 x 1.47 = 68.9kJ
Calculate the amount of heat needed to decompose 25.0g of ammonia NH3 into its elements?
_____Thermo _____2NH3 ---------------> N2 + 3H2_______dH = + 46.19kJ/mole NH3 moles NH3 = 25.0 g / MW NH3 g/mole NH3 = ?? moles NH3 heat needed = ?? moles NH3 * 46.19kJ/mole NH3 = ?? kJ
Calculate the amount of heat released when 2.00 gram of NH3 is combusted.Hint: How many moles of NH3 is this?
The combustion of ammonia with oxygen occurs by the reaction 4 NH3(g) + 7 O2(g) => 4 NO2(g) + 6 H2O(g) Delta H = –1132 kJ Plz explain the process of soving.... a) 33.2 kJ b) 87.0 kJ c) 11.4 kJ d) 16.6 kJ
How do I calculate the amount of water produced by the combustion of 16 g of methane?
To solve such questions you must write a balanced reaction first.CH4 (g)+2O2 (g)————-> CO2 (g) +2H2O (g)Now according to the reaction 1 mole of CH4 gives 2 moles of H2OAnd 16 grams of Methane mean 1 mole (moles=given mass/molecular mass. Molecular mass of methane is 16)so 1 mole of methane = 16/16 (given mass 16 and molecular mass 16)since 1 mole of methane is getting used up to form 2 mole of water.number of moles=given mass/molecular mass ——- eq. 1molecular mass of water is 18.2=x/18 ( where x is the mass of water produced , using equation 1)2*18=36gramsso 36 grams of water will be produced.
How many moles of oxygen O2 are needed for a complete combustion of two moles of butane?
Wait… I see it before… Why?You have to derive the chemical equation for the combustion of butane (C4H10)The chemical equation must be something like C4H10 + O2 → CO2 +H2O (without the coefficients)I don’t know whether you know how to balance the equation, but after balancing the equation, it should be something like this:C4H10 + 6.5 O2 → 4 CO2 + 5 H2OEach mole of butane needs 6.5 moles of oxygen, so 13 moles of oxygen is required for 2 moles of butane in a complete combustion.This is copied and pasted from a comment I gave in Trevor Cheung's answer to Element P has an electronic configuration of 2,8,6 element R has an electronic configuration of 2,8,8,1.What is likely to form if P and R combine?
Calculate the enthalpy of the reaction: 2NO(g) + O2(g) --> 2NO2(g)?
Calculate the enthalpy of the reaction: 2NO(g) + O2(g) --> 2NO2(g) given the following reactions and enthalpies of formation: 1/2 N2(g) + O2(g) --> NO2(g), delta H *A = 33.2 kJ 1/2 N2(g) + 1/2O2(g) --> NO(g), delta H *B= 90.2 kJ Calculate the enthalpy of the reaction: 4B(s)+3O2(g) --> 2B2O3(s) given the following pertinent information: B2O3(s) + 3H2O(g) --> 3O2(g) + B2H6 (g) delta H *A = +2035 kJ 2B(s) + 3H2(g) --> B2H6(g) delta H *B = +36 kJ H2(g) + 1/2O2(g) --> H2O(l) delta H *C = -285 kJ H2O(l) --> H2O(g) delta H *D = +44 kJ Can someone show me step by step? I would be very grateful because I do not understand Hess's law very well.
Ammonium nitrate can decompose explosively when heated according to the equation below. 2 NH4NO3(s) 2 N2(g) +?
find moles. using molar mass 470 grams of NH4NO3 @ 1 mol / 80.04 grams NH4NO3 = 5.872 moles NH4NO3 by the equation: 2 NH4NO3(s) --> 2 N2(g) + 4 H2O(g) +O2(g) 5.872 moles NH4NO3 produces 7/2 as many moles of gases = 20.55 moles of gases find litres PV = nRT (1.00 atm)(V) = 20.55 mol (0.08206 L-atm/mol-K)(698 K) volume of gas produced is = 1177 litres if the 470 grams was actually written with a decimal given, as "470." then you have a 3 sig fig problem & your answer is rounded to 1180 Litres if the 470 grams was actually written without a decimal given, as "470" then you have a 2 sig fig problem & your answer is rounded to 1200 Litres (trailing zeros are not significant unless a decimal point is written) =======================================... p.s. i am editting in a note regarding your question about mole fraction of Ar vs. N2 I answered it too late, & so will be deleting my answer. but your answerer gave you that answere in the form of %, & not in the form you had requested, "mole fraction" here is the answer which I had & did delete: at STP there is 22.4 litres / mole so \find the average molar mass: (1.549 g/L) ( 22.4 litres / mole) = 34.70 grams per mole if X is the mole fraction of Ar then (1-X) is the mole fraction of N2 use mole fractions to get that average molar mass of 34.7 g/mol: (X mol Ar) (39.95 g/mol) & (1 - X mol N2) (28.01 g/mol) = 34.7 39.95X & 28.01 - 28.01X = 34.7 39.95X - 28.01X = 34.7 - 28.01 11.94 X = 6.69 X = 0.5603 is the mole fraction of Ar so 0.4397 is the mole fraction of N2
When hydrogen peroxide decomposes, what mass of hydrogen peroxide is needed in a solution to produce 1.6g of oxygen gas?
The Molar mass of Oxygen is 32 g/molTherefore moles of Oxygen = 1.6/32 =0.052 H2O2 = 2 H2O + O2So, 2 x 0.05 = 0.1 moles of peroxide are needed.The molar mass of peroxide is 34 g/molMass = moles x Mr = 0.1 x 34 = 3.4gAlthough in reality the yield is unlikely to be 100% so a better answer would be “more than 3.4g”
What happens when you combine calcium oxide and water?
When we combine Calcium oxide (CaO) known as lime with water it will form slacked lime >Ca(OH)2CaO + H20 >> Ca (OH)2If you are also suffering from the tension and less focus on study, then you will find out a solution here- How to increase concentration and memory power!!Go to the link and check out. :)