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What is the limiting reactant when 3.41 g of nitrogen react with 2.79g of hydrogen to produce ammonia and how many grams of ammonia are produced?
The reaction for the Haber process is
N2 + 3 H2 ⇌ 2 NH3
Amount of N2 = 3.41/28.0 = 0.122mol
Amount of H2 = 2.79/2.0 = 1.40mol
According to the stoichiometry of the reaction, 1 mol of N2 reacts with 3 mol of H2. 0.122mol of N2 will therefore react with only 0.366mol of H2, but there is 1.40mol of H2 available. Thus H2 is in excess and N2 is the limiting reactant.
1mol of N2 reacts to form 2 mol of NH3.
Under the maximum possible yield, 0.122mol of N2 reacts to form 0.244mol of NH3.
Mass of NH3 = 0.244 x 17.0 = 4.15g
What else can I help you with?
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 an excess of nitrogen is reacted with 3.41 grams of hydrogen how many grams of ammonia can be produced?
Balanced equation first. N2 + 3H2 >> 2NH3 (hydrogen is limiting and drives the reaction ) 3.41 grams H2 (1mol/2.016g )(2mol NH3/3mol H2 )(17.034g NH3/1mol NH3 ) = 19.2 grams of ammonia produced ( this is called the Born-Haber process )
What is the maximum amount in grams of SO3 that can be produced by the reaction of 1.0g of S with 1.0g of O2 via the equation belowS O2-- SO3 not balanced?
To determine the maximum amount of SO3 that can be produced, we need to find the limiting reactant first. Given 1.0g of S and 1.0g of O2, we calculate the number of moles for each reactant. Then, we find the mole ratio from the unbalanced equation and determine which reactant is limiting. Finally, we can calculate the maximum amount of SO3 that can be produced from the limiting reactant.
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.
Why are the amounts of products formed in a reaction determined only by the limiting reactant?
The limiting reactant is the reactant that is completely consumed first, limiting the amount of products that can be formed. Once the limiting reactant is used up, the reaction stops, regardless of the amounts of excess reactants present. This results in the amounts of products formed being determined solely by the limiting reactant.
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.
What mass of ammonia NH3 could be produced from the reaction of 44.5 g of nitrogen N2 and 2.58 g of hydrogen H2?
The equation for the reaction is N2 + 3 H2 -> 2 NH3. The gram atomic mass of nitrogen is 14.0067, and the gram atomic mass of hydrogen is 1.00794. Therefore, the mass fraction of nitrogen in ammonia is 14.0067/[14.0067 + (3*)(1.00794)] or about 0.8225, and, since nitrogen and hydrogen are the only two elements present, the mass fraction of hydrogen is 1*- 0.8225 or about 0.1775. The mass fraction of nitrogen in the amounts of nitrogen and hydrogen specified is 44.5/(44.5 + 2.58) or about 0.945. Therefore, hydrogen is the limiting reactant in this mixture, and the mass of ammonia produced is 2.58/0.1775 or 14.5 grams, to the justified number of significant digits. ________________ *An exact integer.
Ammonia gas NH3 can be manufactured by combining hydrogen and nitrogen gases If 28 g of nitrogen and 7 g of hydrogen are available how many grams of ammonia can be produced?
The balanced chemical equation for the formation of ammonia from nitrogen and hydrogen is N2 + 3H2 → 2NH3. From the equation, it can be seen that 1 mole of nitrogen reacts with 3 moles of hydrogen to produce 2 moles of ammonia. Calculate the moles of nitrogen and hydrogen provided, determine the limiting reactant, and then use stoichiometry to find the grams of ammonia that can be produced.
When 28 g of nitrogen and 6 g of hydrogen react 34 g of ammonia are produced. If 80 g of nitrogen react with 4 g of hydrogen how much ammonia will be produced?
Using the law of multiple proportions, we can see that the ratio of nitrogen to hydrogen in ammonia is 28:6 = 4.67:1. Therefore, for 80 g of nitrogen, 80/4.67 = 17.12 g of hydrogen would be needed to react completely. Since only 4 g of hydrogen is available, the limiting reactant is hydrogen and only 6 g of ammonia will be produced.
What is the limiting reagent when 150.0g of nitrogen react with 32.1 g of hydrogen?
To determine the limiting reagent, calculate the moles of each reactant: 150.0g nitrogen is 5.36 moles and 32.1g hydrogen is 31.8 moles. Using the balanced chemical equation, you can see that nitrogen is the limiting reagent because it will be completely consumed before all the hydrogen is reacted.
If an excess of nitrogen is reacted with 3.41 grams of hydrogen how many grams of ammonia can be produced?
Balanced equation first. N2 + 3H2 >> 2NH3 (hydrogen is limiting and drives the reaction ) 3.41 grams H2 (1mol/2.016g )(2mol NH3/3mol H2 )(17.034g NH3/1mol NH3 ) = 19.2 grams of ammonia produced ( this is called the Born-Haber process )
Why is the term limiting used to describe the limiting reactant?
The term "limiting" is used to describe the reactant that is completely consumed in a chemical reaction, thus limiting the amount of product that can be formed. It determines the maximum amount of product that can be produced based on its stoichiometry and quantity.
What is the reactant called that gets used up first?
The Limiting Reactant is the reactant that runs out first in a reaction.
Why does the amount of product formed is directly proportional to the amount of limiting reactant used?
The amount of product formed is directly proportional to the amount of limiting reactant used because the limiting reactant determines the maximum amount of product that can be produced in a chemical reaction. Any excess reactant beyond the limiting reactant will not contribute to the formation of additional product. Thus, the amount of product formed is dictated by the amount of limiting reactant available.
How do you find out how many moles of lead will be produced from a limiting reactant?
Knowing the limiting reactant, ignore other reactants and calculate the product (lead) based on just that one reactant using the coefficients of the balanced equation.
What is the definetions of limiting reactant in chemistery?
In a chemical reaction the limiting reactant is the reactant that there is the least of in the reaction; it determines the amount of product formed. In a chemical reaction it is the reactant that gets completely "used up"
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.
