Mass of 4 mole CO2 , this is 4 * (12 + 2*16) = 176 grams
The reaction equation gives the RATIO of moles reactant to moles product. With known molar mass the (mass) yield can be calculated. (Of course other reactants are to be in excess! for complete reaction of the reactant involved for this yield)
The Limiting Reactant is the smaller number once you compare the two reactants with one product. The product that you are comparing them both with must be the same. The Excess Reactant is the larger number, or the amount left over in the chemical reaction.
The amount of mass given off as gas during a chemical reaction depends on the stoichiometry of the reaction and the molar mass of the gas produced. It can be calculated using the ideal gas law, which relates the pressure, volume, temperature, and number of moles of a gas.
To determine the mass of silver chloride produced, we need to know the balanced chemical equation for the reaction between silver nitrate (AgNO3) and sodium chloride (NaCl) that produces silver chloride (AgCl) as a precipitate. Once we have the balanced equation, we can use the stoichiometry of the reaction to determine the number of moles of AgCl produced, and then convert that to mass using the molar mass of AgCl.
The solving of mole to mole problems through stoichiometry is based in ratios. Assuming the reaction formulas are simplified and balanced the moles consumed and produced can be calculated. For example in the simple combustion reaction of ethane and oxygen of:CH4 + 2O2 = CO2 +2H20 we see a molar ratio of 1:2 between methane combusted and waters produced.
The equation is 2C4H10 + 13O2 --> 8CO2 + 10H2O This means that for each mole of butane there are 5 moles of water produced. We have 7.01 g of butane = 7.01/58 moles of butane = 0.12 moles. Thus we will get 5 x 0.12 moles of water, = 5 x 0.12 x 18 g of water = 10.88 g.
The reaction equation gives the RATIO of moles reactant to moles product. With known molar mass the (mass) yield can be calculated. (Of course other reactants are to be in excess! for complete reaction of the reactant involved for this yield)
You gave no starting moles/mass.
find moles: 20.0 grams of Al @ (27.0 g/mol) = 0.7407 moles of Al by the reaction: 2 moles Al+3CuSO4 → Al2(SO4)3 +3 moles Cu 0.7407 moles of Al produces 3/2 's as many moles of Cu = 1.11 moles of Cu find mass, using molar mass: 1.11 moles of Cu @ (63.5 g/mol) = 70.6 grams of Cu your answer is 70.6 g
Balanced chemical equation along with the stoichiometric ratios derived from that chemical reaction. A + B --> 2C mass of A * 1/molar mass of A = moles of A Moles of A * 2 moles of C/mole of A = moles of C Moles of C * molar mass of C = mass of C Also, you must think about limiting reagents, because if there is not enought reactant B to react with the amount of reactant A then the amount of reactant B will limit the production of product C!
The Limiting Reactant is the smaller number once you compare the two reactants with one product. The product that you are comparing them both with must be the same. The Excess Reactant is the larger number, or the amount left over in the chemical reaction.
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.
the energy produced by the reaction.
The amount of mass given off as gas during a chemical reaction depends on the stoichiometry of the reaction and the molar mass of the gas produced. It can be calculated using the ideal gas law, which relates the pressure, volume, temperature, and number of moles of a gas.
The mass of heavy water produced when 7,00grams of oxygen reacts with excess D2 is 7,875 g.
Mg + 2HCl => MgCl2 + H2 2 moles HCl : 1 mole H2. 2.02 x .5 moles = 1.01 1.01g H2
The molecular mass of iron III nitrate is 241.86g/mol. Mass divided by molecular weight gives moles, which is 0.02481 moles (6/241.86)