3.03
It measures the amount of reactants actually produced in a reaction compared to the amount that would theoretically be produced if 100% of the reactants were converted to products according to the stoichiometry of the reaction. It is found by: actual moles of products ÷ predicted moles of products * 100%
3:4 or 75%
It is a blue print or a copy on paper for the specifications of an item to be produced or manufactured
You can only calculate the empirical formula because you do not have a mass of this compound given. To do the empirical formula assume 100 grams and change percent to grams. Get moles. 80 grams Carbon (1 mole C/12.01 grams) = 6.66 moles C 20 grams hydrogen (1 mole H/1.008 grams) = 19.84 moles H the smallest becomes 1 in the empirical formula and the other number is divided by it, Thus; H/C 19.84 moles H/6.66 moles C = 2.9, which we call 3 so, CH3 --------------- is the empirical formula To get the molecular formula tour question needed to read; How to calculate molecular formula from such ans such mass of compound with these percentages of elements, Which, of course, your question did not provide. Then you would have divided that given mass by the mass total of the elements of the empirical formula, got a whole number by which you would have multiplied the numbers of your empirical formula to get molecular formula.
About 1.2 times ten to the 24th power. This is so much more money than has ever existed that it's almost meaningless. (To give some perspective, there's about 1.2 times ten to the 12th dollars in circulation right now; if you piled all of it up, it would take a trillion such piles to make two moles of dollars.)
Balanced equation. N2 + 3H2 --> 2NH3 1.4 moles H2 (2 moles NH3/3 moles H2) = 0.93 moles NH3 produced =======================
For every 1 mole of CH4 that reacts, 1 mole of CO2 is produced. Therefore, 4 moles of CH4 will produce 4 moles of CO2. To calculate the mass of CO2 produced, you would need to multiply the moles of CO2 by its molar mass (44 g/mole) to get the total mass produced.
For every mole of propane (C3H8) that reacts, 4 moles of water (H2O) are produced. Since 1 mole of gas occupies about 22.4 L at STP, if 4.50 L of propane reacts, you would need to convert the volume of propane to moles using the ideal gas law to determine the moles of water produced.
To determine the mass of ammonia produced, you first need to calculate the moles of hydrogen gas present. Then, you can use the stoichiometry of the balanced chemical equation for the reaction between hydrogen and nitrogen to find the moles of ammonia produced. Finally, using the molar mass of ammonia, you can convert moles to grams to find the mass of ammonia produced.
Mg3N2 forms in insufficient air because magnesium reacts with nitrogen gas to form magnesium nitride. When there is not enough oxygen present, magnesium is more likely to react with nitrogen to form Mg3N2 rather than magnesium oxide (MgO).
The balanced chemical equation for the reaction would be: 4KO2 + 2CO2 → 2K2CO3 + 3O2 Since the molar ratio between KO2 and O2 is 4:3, 0.400 mol of KO2 would produce: 0.400 mol KO2 * (3 mol O2 / 4 mol KO2) = 0.300 mol O2
The chemical formula of magnesium nitride would be Mg3N2.
what is ag2o
2KClO3 --> 2KCl + 3O2For every 3 moles of oxygen gas produced, 2 moles of potassium chlorate are used.6 moles O2 * (2 moles KClO3 reacted / 3 moles O2 produced) = 4 moles KClO3
1 mole of magnesium reacts with 2 moles of nitric acid to produce 1 mole of magnesium nitrate. Therefore, if 1 mole of magnesium nitrate is produced from 2 moles of nitric acid, 8 moles of nitric acid would produce 4 moles of magnesium nitrate.
For every mole of oxygen consumed in the reaction 2H2 + O2 -> 2H2O, two moles of water are produced. Therefore, if 0.633 moles of oxygen are consumed, the number of moles of water produced would be 2 x 0.633 = 1.266 moles.
The balanced chemical equation for the reaction is 2SO2 + O2 -> 2SO3. This means that for every 1 mole of O2 consumed, 2 moles of SO3 are produced. Therefore, 1.32 moles of O2 would produce 2.64 moles of SO3.