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If 25.0 grams of potassium chlorate decomposes, how many grams of potassium chloride will be produced What is the limiting reactant?
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What is the limiting reagent when sodium hydroxide and copper chloride react?
The limiting reagent is the reactant that is completely consumed first in a chemical reaction. To determine the limiting reagent in the reaction between sodium hydroxide and copper chloride, you would need to compare the moles of each reactant present and see which one is in excess and which one is limiting.
How many grams of silver chloride are formed when 10.0 g of silver nitrate reacts with 15.0 g of barium chloride?
To find the limiting reactant, we need to determine how many grams of silver chloride can be produced from each reactant and compare the results. Calculate the amount of silver chloride that can be produced from 10.0 g of silver nitrate. Calculate the amount of silver chloride that can be produced from 15.0 g of barium chloride. The reactant that produces the lesser amount of silver chloride will be the limiting reactant.
Which reactant in the reaction of sodium carbonate and calcium chloride dihydrate is the limiting reactant if 1.00 gram of each reagent is used?
To determine the limiting reactant, calculate the moles of each reactant using their molar masses. Then, use the stoichiometry of the reaction to determine which reactant will be consumed first. Whichever reactant produces the lesser amount of product will be the limiting reactant.
What salt is formed when potassium hydroxide reacts with suphuric acid?
The salt formed by potassium hydroxide and sulphuric acid is potassium sulphate (K2SO4). Though if potassium hydroxide is the limiting reagent potassium bisulphate (KHSO4) will also form.
How many grams of aluminum chloride could be produced from 34.0 grams of aluminum and 39.0 grams of chlorine gas?
To find the limiting reactant, we need to calculate the moles of each reactant. Then, use the stoichiometry of the balanced chemical equation to determine which reactant limits the amount of aluminum chloride that can be produced. Finally, calculate the mass of aluminum chloride produced based on the limiting reactant.
How many moles of potassium bromide can be produced from the reaction of 2.92 moles of potassium with 1.78 moles of bromine gas?
First write a balanced chemical equation: 2K + Br2 ---> 2KBR Find the limiting reactant by using the moles of each element and determining which one gives you the smallest number of moles of potassium bromide. 2.92 mol K (2 mol KBr/2 mol K)= 2.92 mol KBr 1.78 mol Br2 (2 mol KBR/1 mol Br2)=3.56 mol KBr potassium is your limiting reactant so the max. number of moles of KBr that can be produced is 2.92 mol of KBr
Cl2(g) plus 2Kl(aq)-- and gt I2(s) plus 2KCl(aq) Calculate the mass of iodide produced when 4.50x1000g of chloride gas is bubbled through an escess amount of potassium iodide solution?
First, determine the limiting reactant by using the stoichiometry of the reaction. Then calculate the amount of iodine produced based on the limiting reactant. Finally, convert the amount of iodine produced to mass using its molar mass.
If you had excess aluminum how many moles of aluminum chloride could be produced from 31.0 g of chlorine gas Cl2?
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.
How the amount of potassium iodide added to the potassium iodate (V) solution in the conical flask affects the amount of iodine liberatedexplain.?
If more potassium iodide is added to the potassium iodate (V) solution in the conical flask, there will be more iodine liberated. This is because potassium iodide reacts with potassium iodate (V) to produce iodine. Therefore, increasing the amount of potassium iodide increases the rate of reaction and the amount of iodine generated.
How many grams of calcium chloride will be produced when 30.0g of calcium carbonate are combined with 12.0g of hydrochloric acid?
Based on the chemical reaction: CaCO3 + 2HCl -> CaCl2 + CO2 + H2O Calculate the limiting reactant by converting each reactant to moles and comparing the ratios. The limiting reactant is calcium carbonate. Calculate the theoretical yield of calcium chloride using the stoichiometry of the reaction.
What mass of sodium chloride is formed if 22.99 grams of sodium combines with 35.54 grams of chlorine?
The molar mass of sodium chloride is 58.44 g/mol. To find the mass of sodium chloride formed, you need to compare the moles of sodium and chlorine to determine the limiting reactant. Calculate moles of sodium and chlorine, determine limiting reactant, and use stoichiometry to find mass of sodium chloride formed.
Does the amount of potassium iodide added to the potassium iodate solution affect the amounts of iodine liberated in iodometric titration?
Yes, the amount of potassium iodide added to the potassium iodate solution in iodometric titration affects the amount of iodine liberated. Potassium iodide serves as a reducing agent, reacting with the iodate ion to form iodine. The quantity of potassium iodide added determines the rate and completeness of this reaction, impacting the amount of liberated iodine available for titration.
