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The balanced chemical equation for the reaction between calcium and chlorine gas to produce calcium chloride is: Ca + Cl2 -> CaCl2. From this equation, we can see that one mole of calcium reacts with one mole of chlorine gas to produce one mole of calcium chloride.
The molar mass of calcium is 40.08 g/mol and the molar mass of chlorine gas is 70.90 g/mol. This means that 10.0 grams of calcium is equivalent to 0.249 moles of calcium and 20.0 grams of chlorine gas is equivalent to 0.282 moles of chlorine gas.
Since the ratio of calcium to chlorine gas in the balanced chemical equation is 1:1, this means that 0.249 moles of calcium would react completely with 0.249 moles of chlorine gas, leaving an excess of 0.033 moles (or 2.34 grams) of chlorine gas.
The limiting reactant in this reaction is calcium, and the maximum amount of calcium chloride that can be produced is equivalent to the number of moles of the limiting reactant, which is 0.249 moles (or 27.8 grams) of calcium chloride.
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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.
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 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.
1. When 200.2g of sulfur reacts with 100.3g of chlorine to produce disulfur dichloride acts as the limiting reactant?
The limiting reactant is chlorine.
You are given 34.0 g of aluminum and 39.0 g of chlorine gas. If you had excess chlorine how many moles of aluminum chloride could be produced from 34.0 g of aluminum?
First, calculate the number of moles of aluminum in 34.0 g using its molar mass. Then, determine the limiting reactant by converting the moles of aluminum to moles of aluminum chloride using the mole ratio from the balanced chemical equation. Finally, calculate the moles of aluminum chloride that can be produced based on the limiting reactant.
What is the maximum mass of aluminum chloride that can be formed when reacting 28.0g of aluminum with 33.0g of chlorine?
The formula of aluminium chloride is AlCl3. The atomic weight of aluminium is 27 and that of chlorine is 35.5. That means 35.5*3 grams of chlorine will combine with 27 grams of aluminium. So 33 grams of chlorine will combine with 8.37 grams of aluminium. The addition of both makes it 41.37 grams. In this reaction, the whole chlorine will be utilized and only part of the aluminium.
How many grams of aluminum chloride could be produced from 35.5g of aluminum and 39.0g of chlorine gas?
The balanced chemical equation for the reaction is: 2Al + 3Cl₂ → 2AlCl₃ Calculate the limiting reactant: Moles of Al: 35.5g / molar mass of Al Moles of Cl₂: 39.0g / molar mass of Cl₂ Determine which reactant gives the lower amount of AlCl₃ produced. Once you know the limiting reactant, use stoichiometry to calculate the grams of AlCl₃ produced.
What is the limiting reactant when 3.14g of nitrogen react with 2.79g of hydrogen to produce ammonia and how many grams of ammonia are produce?
To determine the limiting reactant, we need to find the moles of each reactant. Then, we calculate the amount of ammonia that can be produced from each reactant. Whichever reactant produces the least amount of ammonia is the limiting reactant. Finally, we calculate the grams of ammonia produced based on the limiting reactant.
If 25.0 grams of potassium chlorate decomposes, how many grams of potassium chloride will be produced What is the limiting reactant?
I Don't knows Sorry
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 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.
How many moles of hydrogen chloride can be produced from 0.490 grams of Hydrogen and 50.0 grams of chlorine?
The atomic mass of hydrogen is 1.008 and that for chlorine is 35.45. The moles of hydrogen available are therefore 0.490/1.008 = 0.486 and the moles of chlorine available, 50/35.45, are greater than 1. Each molecule of hydrogen chloride requires one atom each of chlorine and hydrogen. Therefore, with the specified conditions, hydrogen is stoichiometrically limiting, and 0.486 moles of HCl can be made.
