Since the product, K2SO4 has twice as many potassium atoms as the reactant, KNO3, we may conclude that it takes two moles of reactant to produce one mole of product. The sulfate and nitrate ions will not be the limiting factor.
To determine the limiting reactant and thus how much X remains, first calculate the moles of Z produced using the mole ratio from the balanced equation, given moles of X and Y. Then, identify the limiting reactant by comparing the moles of Z produced from each reactant. Finally, determine the moles of X remaining by subtracting the moles of Z produced from the moles of X initially given.
Approximately 2.5 moles of ATP are produced from each mole of NADH oxidized in the electron transport chain (ETC), while about 1.5 moles of ATP are produced from each mole of FADH2 oxidized in the ETC. This difference arises from the different redox potentials and positions of NADH and FADH2 in the electron transport chain.
That would be 2 moles of OH- (hydroxide) ions because there are 2 of them associated with each 1 of the group II metal.
Each molecule of C6H6 contains 6 carbon atoms, so when 1 mole of C6H6 decomposes, 6 moles of carbon atoms are obtained. Therefore, in a 1.68 mole sample of C6H6, 6 × 1.68 = 10.08 moles of carbon atoms can be obtained from the decomposition.
Boron trifluoride is BF3. So each mole of BF3 contains 1 moles of boron (B) and 3 moles of fluorine (F). Thus, 3 moles of BF3 contains NINE moles of fluorine.
There are 5.2 moles of ions in 2.6 moles of K2SO4. This is because each formula unit of K2SO4 dissociates into three ions: 2K+ ions and 1 SO4^2- ion. Thus, for every mole of K2SO4, you have 3 moles of ions.
To find the number of moles of K2SO4 in 15.0 grams, first calculate the molar mass of K2SO4 (K: 39.10 g/mol, S: 32.07 g/mol, O: 16.00 g/mol). Molar mass of K2SO4 = 2(39.10) + 32.07 + 4(16.00) = 174.26 g/mol Now, divide the given mass by the molar mass to find the number of moles: 15.0 g / 174.26 g/mol = 0.086 moles of K2SO4
The ratio between K3PO4 and Al(NO3)3 is 1 : 1 We have 2 moles of each reactant so we have not a reactant in excess The ratio between K3PO4 and KNO3 is 1 : 3 Moles KNO3 produced = 2 x 3 / 1 = 6
A fair question and some tricky points. Firstly, K2SO4 can be seen as 2K+[SO4]. If you put it into water, you get 2 potassium ions for each K2SO4. That's important. The rest is just wrestling with the numbers. We have 200cm3 of 2.0M K2SO4. How many moles is that? Well we'd have 2 moles in 1000cm3, so... (1000/200 = 5) we have 5 times less volume meaning 5 times fewer moles (2.0/5 = 0.4) so we have 0.4 moles K2SO4. Good. Now we put that into 800cm3 of water. There's a trick, here! It's not 0.4 moles into 800cm3 water, because our 0.4moles are ALREADY in their own 200cm3 of water. So remember to add the volumes up, too. But this just makes things so much easier for us! Because we end up with 1 litre which is the volume component of the molarity unit. (M = g/dm3 or g/L, same thing) So it's 0.4 moles of K2SO4 in 1litre of water, meaning 0.4M K2SO4. Finally....we want K+ ions, not K2SO4. Remember we had two K+ ions for every K2SO4? 2 x 0.4 = 0.8. So our answer is: we have 0.8M K+ solution in the end.
To determine which substance contains the largest number of moles, you need to divide the given mass of each substance by its molar mass to find the number of moles. The substance with the largest number of moles will have the smallest molar mass. In this case, Neon (Ne) has the smallest molar mass, so it contains the largest number of moles.
To determine the limiting reactant and thus how much X remains, first calculate the moles of Z produced using the mole ratio from the balanced equation, given moles of X and Y. Then, identify the limiting reactant by comparing the moles of Z produced from each reactant. Finally, determine the moles of X remaining by subtracting the moles of Z produced from the moles of X initially given.
potassium nitrate
Approximately 2.5 moles of ATP are produced from each mole of NADH oxidized in the electron transport chain (ETC), while about 1.5 moles of ATP are produced from each mole of FADH2 oxidized in the ETC. This difference arises from the different redox potentials and positions of NADH and FADH2 in the electron transport chain.
To calculate the moles of aluminum chloride produced, you would first need to determine the limiting reactant. Compare the moles of each reactant (Aluminum and Cl2) using their molar masses. Whichever reactant produces fewer moles of aluminum chloride would be the limiting reactant. Once you have that, you can use the stoichiometry of the balanced chemical equation to calculate the moles of aluminum chloride produced.
One mole of glucose yields 36 moles of ATP through cellular respiration, while one mole of triglyceride (composed of lauric acid) yields 129 moles of ATP through beta oxidation and the citric acid cycle. Therefore, with 3.5 moles of each, the triglyceride will produce more energy by yielding 451.5 moles of ATP, compared to the 126 moles produced by the glucose.
In each mole of ethanol there are two moles of carbon atoms. One mole of carbon atoms is required to form one mole of carbon dioxide, so two moles will form two moles. Thus each mole of ethanol completely combusted will form two moles of carbon dioxide.
To find the limiting reactant, first convert the masses of Al and Cl2 to moles. The balanced equation shows that 2 moles of Al reacts with 3 moles of Cl2 to produce 2 moles of AlCl3. Determine the moles produced by each reactant and find the limiting reactant. Finally, calculate the mass of AlCl3 produced using the limiting reactant.