The equation for the reaction is 2 C2H2 + 5 O2 -> 4 CO2 + 2 H2O. Therefore, the minimum mole ratio of O2 to C2H2 to permit complete reaction is 5/2. The actual ratio present, 7.40/2.40 or about 3.08, is greater than this minimum; therefore, all the available C2H2 will react. The reaction equation shows that the number of moles of H2O produced is the same as the number of moles of C2H2 reacted. Therefore, 2.40 moles of H2O are produced. The gram molecular mass of H2O is about 18.015; therefore the mass of water produced is 43.2 grams, to the justified number of significant digits.
26.20
62.1 - 62.5
Balanced equation first. N2 + 3H2 >> 2NH3 (hydrogen is limiting and drives the reaction ) 3.41 grams H2 (1mol/2.016g )(2mol NH3/3mol H2 )(17.034g NH3/1mol NH3 ) = 19.2 grams of ammonia produced ( this is called the Born-Haber process )
i bet its 2.0g
You can't answer this question unless you know the the NO2 was formed FROM. You need to write the balanced reaction for the reaction and then use stoichiometry to solve for the amount of oxygen produce.See the Related Questions to the left for how to write a balanced reaction and how to use stoichiometry to solve this type of problem.
4
26.20
The mass of nitrous oxide is 262,8 g.
Nothing is produced, 500g potassium chlorate will be the same 500 g potassium chlorate after reaction. Actually there is no reaction at all.
Following the Law of Conservation of Mass (see link below), there will be 20 grams of products in a reaction of 20 grams of reactions.
Two moles of water are produced.
Depends on what the last reaction is. And probably what the first reaction was too. Insufficient information in the question.
Answer= 4 mol H3PO4 - White Plains Public Schools
To calculate the grams of dichloromethane produced, we first need to convert the mass of methane from kilograms to grams, which is 1,540 grams. Given a yield of 48.2%, we can multiply this by the yield percentage to find the actual amount of dichloromethane produced: 1,540 grams of methane x 0.482 = 742.28 grams of dichloromethane.
51.4 - 51.8
c. 16.0 grams CH4/1 mol C04
322 grams.