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What does the percent yield of a reaction measure?
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%
What is limiting reactant in 2Al plus 6HCl equals 2AlCl3 plus 3 H2?
2Al(s) + 6HCl(aq) = 2AlCl3(aq) + 3H2(g). The molar ratios are 2:6 :: 2:3 This means that two moles of aluminium will react to completion with exactly six moles of hydrocchloric acid. At completion there will be two moles of aluminium chloride and three moles of hydrogen gas. So if you had 8 moles of hydrochloric acid and only 2 moles of aluminium, then the aluminium would be the limiting reactant, because from the 8 moles hydrochloric acid you are using only six moles, leaving two moies on hydrochloric acid unreacted. Conversely, if you had 3 moles of Al and 6 moles HCl , then the HCl would be the limiting reactant. Because the 6 moles of acid would only react with two moles of Al leaving one mole Al unreacted.
What volumen of 0.100 MknO4 would be requiere totitrate 0.33g k2cuc2042.2Ho?
To determine the volume of 0.100 M KMnO4 required to titrate 0.33 g of K2Cu(C2O4)·2H2O, first calculate the moles of K2Cu(C2O4)·2H2O. The molar mass of K2Cu(C2O4)·2H2O is approximately 197.12 g/mol, so 0.33 g corresponds to about 0.00167 moles. The reaction between KMnO4 and K2Cu(C2O4) typically involves a 1:1 stoichiometry, which means you would need 0.00167 moles of KMnO4. Using the molarity of KMnO4 (0.100 M), the volume required would be approximately 16.7 mL (calculated using the formula: Volume = Moles / Molarity).
What would you need to do to calculate the molality of 0.2 kg of NaCI in 3 kg of?
To calculate the molality of 0.2 kg of NaCl in 3 kg of water, first determine the number of moles of NaCl by using its molar mass (approximately 58.44 g/mol). Convert 0.2 kg of NaCl to grams (200 g) and calculate the moles: ( \text{moles} = \frac{200 \text{ g}}{58.44 \text{ g/mol}} ). Finally, use the formula for molality, which is moles of solute per kilogram of solvent: ( \text{molality} = \frac{\text{moles of NaCl}}{3 \text{ kg of water}} ).
What would 0.032 moles of balls in a ball pit be explain math?
To find the number of balls in a ball pit containing 0.032 moles of balls, you can use Avogadro's number, which is approximately (6.022 \times 10^{23}) particles per mole. Multiplying 0.032 moles by Avogadro's number gives: [ 0.032 , \text{moles} \times 6.022 \times 10^{23} , \text{balls/mole} \approx 1.93 \times 10^{22} , \text{balls} ] Therefore, there are approximately (1.93 \times 10^{22}) balls in the ball pit.
How many moles of NH3 are produced when 1.4 moles H2 reacts?
Balanced equation. N2 + 3H2 --> 2NH3 1.4 moles H2 (2 moles NH3/3 moles H2) = 0.93 moles NH3 produced =======================
What mass of CO2 is produced in the reaction of 4 moles of CH4 in excess O2?
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.
C3H8 plus 5O2---3CO2 plus 4H2O. How many moles of water form when 4.50 L propane gas C3H8 completely react?
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.
Why does Mg3N2 form in insufficient air?
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).
If 250.3 L of hydrogen gas (d 0.0899gL ) reacts with an excess of nitrogen gas what mass of ammonia would be 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.
Formula for magnisium and nitrogen?
The chemical formula of magnesium nitride would be Mg3N2.
How many moles of O2 are produced when 0.400 mol of KO2 reacts in this fashion?
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
If 2Ag2O and 4Ag o2 then if you started with 3.25 moles of Ag2O how many moles of O2 would be produced?
what is ag2o
How many moles of potassium chlorate must be used to produce 6 moles of oxygen gas?
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
2 H2 plus O2 2 H2O - How many moles of water will be produced from a fuel cell that consumes 0.633 moles of oxygen?
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.
How many moles of magnesium nitrate are produced when magnesium reacts with 8 moles of nitric acid?
Mg2+(s) + 2HNO3(l)= Mg(NO3)2(aq) + H2(g) since the only mole value given is 8 I must assume this is the limiting reactant. Because of the 2:1 ratio of Nitric acid to Magnesium Nitrate, meaning there must be 2 moles Nitric acid for every 1 mole Magnesium Nitrate formed, 4 moles of Magnesium nitrate will be formed.
How many moles of SO3 will be produced by 1.32moles of O2?
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.
