10 grams NO2 (1 mole NO2/46.01 grams)
= 0.217 moles nitrogen dioxide
======================
95,474 moles
Here are some gas stoichiometry practice problems for you to work on: How many liters of oxygen gas are needed to completely react with 5.0 liters of hydrogen gas to produce water vapor? If 2.0 moles of methane gas react with excess oxygen gas, how many moles of carbon dioxide gas will be produced? A reaction produces 4.0 liters of nitrogen gas. If the reaction is 80 efficient, how many liters of nitrogen gas would be produced if the reaction was 100 efficient? These problems will help you practice applying gas stoichiometry concepts.
It is important to know that the percent of nitrogen in 4.444 moles of ammonium sulfide is the same as the percent of nitrogen in 454 grams or 4843 moles or 96 kg, etc. Remember the law of definite proportions - chemical compounds always contain the same proportion of elements by mass. Perhaps you were asking how much nitrogen is in 4.444 moles of ammonium sulfide given the percent of nitrogen in any given mass. So we'll do that too: find the percent of nitrogen in any given sample and apply it specifically to 4.444 moles.Before we go directly to the 4.444 moles, we have to figure out how much nitrogen is in any amount of ammonium sulfide by percent. To do this, we need the atomic weights of the elements and add them up to find the total molar mass of the compound.Ammonium sulfide = (NH4)2SNitrogen = 14.0 grams × 2 = 28.0 gramsHydrogen = 1.01 grams × 8 = 8.08 gramsSulfur = 32.1 grams------------------------------------------------------Ammonium sulfide = 68.2 gramsNow we take the mass of nitrogen and divide it by the total mass to get our percent.Nitrogen ÷ Ammonium sulfide = % Nitrogen28.0 grams ÷ 68.2 grams = 0.411 = 41.1% Nitrogen in Ammonium sulfideSince we know that in any amount of Ammonium sulfide contains 41.1% of Nitrogen, we can apply it to the mass given.41.1% of 4.444 moles = .411 × 4.444 = 1.83 moles of Nitrogen in 4.444 moles Ammonium sulfide
To calculate the number of moles of sodium borohydride in 100 mg, you need to know the molar mass of the compound, which is 37.83 g/mol. First, convert 100 mg to grams (0.1 g), then divide by the molar mass to get the number of moles, which is approximately 0.0026 moles.
The atomic weight of magnesium is 24.31; therefore, the number of moles in 100 gm is 100/24.31 = 4.11, to the justified number of significant digits.
100/150.158 is 0.666 moles
95,474 moles
I think there is little water or little air?! I'm not 100% sure!
Two atoms of nitrogen form the gaseous, natural state, of nitrogen. 10.62 grams N2 (1 mole N2/28.02 grams)(6.022 X 10^23/1 mole N2)(1 mole N2 atoms/6.022 X 10^23) 0.3790 mole of gaseous nitrogen atoms ------------------------------------------------------ * as you may see, Avogadro's number is over itself as a form of one and is a superfluous step put there for formality's sake
Here are some gas stoichiometry practice problems for you to work on: How many liters of oxygen gas are needed to completely react with 5.0 liters of hydrogen gas to produce water vapor? If 2.0 moles of methane gas react with excess oxygen gas, how many moles of carbon dioxide gas will be produced? A reaction produces 4.0 liters of nitrogen gas. If the reaction is 80 efficient, how many liters of nitrogen gas would be produced if the reaction was 100 efficient? These problems will help you practice applying gas stoichiometry concepts.
It is important to know that the percent of nitrogen in 4.444 moles of ammonium sulfide is the same as the percent of nitrogen in 454 grams or 4843 moles or 96 kg, etc. Remember the law of definite proportions - chemical compounds always contain the same proportion of elements by mass. Perhaps you were asking how much nitrogen is in 4.444 moles of ammonium sulfide given the percent of nitrogen in any given mass. So we'll do that too: find the percent of nitrogen in any given sample and apply it specifically to 4.444 moles.Before we go directly to the 4.444 moles, we have to figure out how much nitrogen is in any amount of ammonium sulfide by percent. To do this, we need the atomic weights of the elements and add them up to find the total molar mass of the compound.Ammonium sulfide = (NH4)2SNitrogen = 14.0 grams × 2 = 28.0 gramsHydrogen = 1.01 grams × 8 = 8.08 gramsSulfur = 32.1 grams------------------------------------------------------Ammonium sulfide = 68.2 gramsNow we take the mass of nitrogen and divide it by the total mass to get our percent.Nitrogen ÷ Ammonium sulfide = % Nitrogen28.0 grams ÷ 68.2 grams = 0.411 = 41.1% Nitrogen in Ammonium sulfideSince we know that in any amount of Ammonium sulfide contains 41.1% of Nitrogen, we can apply it to the mass given.41.1% of 4.444 moles = .411 × 4.444 = 1.83 moles of Nitrogen in 4.444 moles Ammonium sulfide
Ammonium sulfide has the formula (NH4)2S, which contains 2 nitrogen atoms. One mole of (NH4)2S contains 2 moles of nitrogen. In 8.941 mol of (NH4)2S, there are 8.941 x 2 = 17.882 mol of nitrogen. To find the percent of nitrogen by weight, you would compare the molar mass of nitrogen to the molar mass of the compound and then multiply by 100.
To calculate the number of moles of sodium borohydride in 100 mg, you need to know the molar mass of the compound, which is 37.83 g/mol. First, convert 100 mg to grams (0.1 g), then divide by the molar mass to get the number of moles, which is approximately 0.0026 moles.
To find the number of moles of phosphorus atoms in 100 grams of P4S10, we first need to determine the molar mass of P4S10 which is 284.26 g/mol. Next, we calculate the number of moles of P4S10 in 100 grams by dividing 100 g by the molar mass to get 0.352 moles of P4S10. Since there are 4 phosphorus atoms in each P4S10 molecule, there are 0.352 moles x 4 = 1.41 moles of phosphorus atoms in 100 grams of P4S10.
Nitrogen 78% Oxygen 21% Argon 1% Carbon Dioxide 0.5% Air 100%
The atomic weight of magnesium is 24.31; therefore, the number of moles in 100 gm is 100/24.31 = 4.11, to the justified number of significant digits.
atomic weight of carbon dioxide is 2 * 16 + 12 = 44 1 kg = 1000 g 1 kg of co2 has 1000/44 = 22.7 moles yeh i think that's wrong lol isn't it 3.37E25?