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What reactant is limiting if 3000 cm3 of Cl2 at STP react with a solution containing 25 grams of NaBr?
Cl2 + 2NaBr => Br2 + 2NaCl One mole Cl2 reacts with 2 moles NaBr Cl2 = 71 NaBr = 102.9 Molar volume = 22.414 L/mole for ideal gas @STP 3L Cl2 = 3/22.414 = 0.1338 mole 25g NaBr = 25/102.9 = 0.2430 mole 0.1338 moles Cl2 requires 0.2676 moles NaBr for complete reaction The NaBr is the limiting reagent
What is the limiting reactant of 60.0grams of aluminum present when a reaction with 25grams of hydrochloric acid occurs?
To determine the limiting reactant, we need to calculate the amount of each reactant in terms of the same unit (moles). Then, we compare the mole ratio of the reactants from the balanced chemical equation. In this case, convert 60.0 grams of aluminum to moles and 25 grams of hydrochloric acid to moles. Compare the moles of each reactant to determine the limiting reactant.
How many moles of water are formed when 15 mL of 0.40 M LiOH is added to 25 mL of 0.60 M H2SO4?
First, determine the limiting reactant by calculating the number of moles for each reactant (moles = concentration x volume). In this case, LiOH is the limiting reactant since it produces 2 moles of water per 1 mole of LiOH. Therefore, 0.006 moles of LiOH will produce 0.012 moles of water.
0.400mol of octane is allowed to react with 0.800mol of oxygen Which is the limiting reactant?
To determine the limiting reactant, compare the moles of each reactant to the stoichiometry of the balanced. In this case, the balanced equation is: 2 C8H18 + 25 O2 → 16 CO2 + 18 H2O The moles ratio between octane (C8H18) and oxygen (O2) is 2:25. Calculate the ratio for each reactant: Octane: 0.400 mol * (25 mol O2 / 2 mol C8H18) = 5.00 mol O2 needed Oxygen: 0.800 mol O2. Since the actual moles of oxygen available (0.800 mol) are greater than the moles needed for the reaction with octane (5.00 mol), oxygen is in excess and octane is the limiting reactant.
How many grams of water can be produced from 50 grams of hydrogen and 300 grams of oxygen?
2H2 + O2 ===> 2H2O50 grams H2 x 1 mole H2/2 g = 25 moles H2300 grams O2 x 1 mole O2/32 g = 9.375 moles O2Limiting reactant is O2, so maximum moles of H2O formed = 9.375 O2 x 2H2O/1 O2 = 18.75 molesgrams H2O = 18.75 moles H2O x 18 g/mole = 337.5 g H2O = 340 g (to 2 significant figures)
What reactant is limiting if 3000 cm3 of Cl2 at STP react with a solution containing 25 grams of NaBr?
Cl2 + 2NaBr => Br2 + 2NaCl One mole Cl2 reacts with 2 moles NaBr Cl2 = 71 NaBr = 102.9 Molar volume = 22.414 L/mole for ideal gas @STP 3L Cl2 = 3/22.414 = 0.1338 mole 25g NaBr = 25/102.9 = 0.2430 mole 0.1338 moles Cl2 requires 0.2676 moles NaBr for complete reaction The NaBr is the limiting reagent
What is the limiting reactant of 60.0grams of aluminum present when a reaction with 25grams of hydrochloric acid occurs?
To determine the limiting reactant, we need to calculate the amount of each reactant in terms of the same unit (moles). Then, we compare the mole ratio of the reactants from the balanced chemical equation. In this case, convert 60.0 grams of aluminum to moles and 25 grams of hydrochloric acid to moles. Compare the moles of each reactant to determine the limiting reactant.
How many moles of water are formed when 15 mL of 0.40 M LiOH is added to 25 mL of 0.60 M H2SO4?
First, determine the limiting reactant by calculating the number of moles for each reactant (moles = concentration x volume). In this case, LiOH is the limiting reactant since it produces 2 moles of water per 1 mole of LiOH. Therefore, 0.006 moles of LiOH will produce 0.012 moles of water.
0.400mol of octane is allowed to react with 0.800mol of oxygen Which is the limiting reactant?
To determine the limiting reactant, compare the moles of each reactant to the stoichiometry of the balanced. In this case, the balanced equation is: 2 C8H18 + 25 O2 → 16 CO2 + 18 H2O The moles ratio between octane (C8H18) and oxygen (O2) is 2:25. Calculate the ratio for each reactant: Octane: 0.400 mol * (25 mol O2 / 2 mol C8H18) = 5.00 mol O2 needed Oxygen: 0.800 mol O2. Since the actual moles of oxygen available (0.800 mol) are greater than the moles needed for the reaction with octane (5.00 mol), oxygen is in excess and octane is the limiting reactant.
How many grams of water can be produced from 50 grams of hydrogen and 300 grams of oxygen?
2H2 + O2 ===> 2H2O50 grams H2 x 1 mole H2/2 g = 25 moles H2300 grams O2 x 1 mole O2/32 g = 9.375 moles O2Limiting reactant is O2, so maximum moles of H2O formed = 9.375 O2 x 2H2O/1 O2 = 18.75 molesgrams H2O = 18.75 moles H2O x 18 g/mole = 337.5 g H2O = 340 g (to 2 significant figures)
How do you know which reactant in a process is the limiting reactant?
The limiting reactant or reagent can be determined by calculating the number of moles of each reactant/reagent. Whichever is the lowest number of moles is the limiting reagent in the reaction, assuming that stoichiometry is 1;1
What mass of solid agcl is obtained when 25 ml of 0.068m agno3 reacts with excess of aqueous hcl?
By definition, No. of moles = given mass/molecular mass; and also by definition, molar concentration of a solute means the number of moles of solute per liter of solution. Therefore, 25 mL of 0.068 M AgNO3 contains AgNO3 = (0.068 * 25) / 1000 = 0.0017 mol of AgNO3.The equation for the reaction is AgNO3 + HCl -> AgCl + HNO3, showing that 0.0017 mol of AgNO3 gives 0.0017 mol of AgCl. The molecular mass of AgCl = 107+35.5 = 143.5 gTherefore, the mass of AgCl produced by the reacion = No. of moles*molecular mass = .0017*143.5 = 0.24g, to the justified number of significant digits.100 mL of 0.068 M AgNO3 contains AgNO3 = 0.068 molSo,Therefore,
What mass of solid agcl is obtained when 25 ml of 0.068 m agno3 reacts with excess of aqueous hcl?
See it's an easy one..!! AgNO3 + HCl -> AgCl + HNO3 100 mL of 0.068 M AgNO3 contains AgNO3 = 0.068 mol So, 25 mL of 0.068 M AgNO3 contains AgNO3 = (0.068 * 25) / 1000 = 0.0017 mol From the equation, we can see 1 mol of AgNO3 gives 1 mol of AgCl 0.0017 mol of AgNO3 gives 0.0017 mol of AgCl Amount of AgCl can be found this way.! No. of moles = given mass/ molecular mass molecular mass of AgCl = 107+35.5 = 143.5 g Therefore, Given mass = No. of moles*molecular mass = 0.0017*143.5 = 0.244g Note : In your question, you have written 0.068 "m" .. (small) m represents for Molality and (capital) M represents for Molarity..! Hope I helped.. :)
When was Mole's Christmas created?
Mole's Christmas was created on 1994-12-25.
How many molecule in 25 grams of NH3?
25 grams / (17 grams/mole) x 6.022x1023 molecules/mole = 8.9x1023 molecules
How many atoms are there in 20 mole of sulphur?
20 moles x 6.02x10^23 atoms/mole = 1.2x10^25 atoms
How many atoms are in one mole of sucrose?
about 1.4*10^25
