0
What else can I help you with?
What mass of iron is needed to react with 16 grams of sulfur?
iron forms iron sulphide when reacts with sulphur, Fe + S = FeS in this reaction 32 g of sulphur require 56 g of iron so 16 g sulphur requires 28 g of iron.
In the reaction 2 Fe plus 3 S--- Fe2S3 how much iron is required to react with 48.0 g of sulfur?
To determine the amount of iron required, you first need to find the molar mass of sulfur (32.06 g/mol) and the molar mass of iron (55.85 g/mol). Then, use the stoichiometry of the reaction to calculate the amount of iron needed, which is 64.4 g.
What happens to the number of particles at the beginning and end of the iron and sulfur reaction?
In a reaction between iron and sulfur to form iron sulfide, the total number of particles remains the same before and after the reaction. This is due to the law of conservation of mass, which states that matter cannot be created or destroyed in a chemical reaction.
What is the percent of sulfur in iron III sulfate?
It would depend on your definition of percentage. For example, if you were looking for percentage of atom count, you would count the atoms in the compound Fe2(SO4)3. You have 12 Oxygen, 3 Sulfur, and 2 Iron, leaving you with 3 sulfur out of 17 total atoms, or 3/17 or 0.176470588 (about 17.5%). If, on the other hand, you are talking about percentage of mass, then you must calculate the mass of each element in the compound. Iron's atomic mass is 55.847. Sulfur's atomic mass is 32.066. Oxygen's atomic mass is 15.999. Multiply these by their respective ratios. Iron (55.847*2=111.694) Sulfur (32.066*3=96.198) Oxygen (15.999*12=191.988) Add them together to get (111.694+96.198+191.988=399.88) Now, take the overall mass of the sulfur part (96.198) and divide it into the total mass (399.88). The answer is 96.198/399.88 or 0.24056717 or about 24%.
What is the percent composition of a compound that contains 17.6 g of iron and 10.3 g of sulfur has a total mass of 27.9 g compound?
The percent by mass of iron in the stated compound is 100(17.6/27.9) or 63.1 %, to the justified number of significant digits. Since the compound contains only two elements, the percent by mass of sulfur is 100 - 63.1 or 36.9 %. To find the atomic percentages rather than the mass percentages, the mass for each element must be divided by the gram atomic mass of that element. 17.6 grams constitutes 17.6/55.457 or about 0.313 gram atoms of iron, and 10.3 grams of sulfur constitutes 10.3/32.06 or about 0.321 gram atoms of sulfur. The atomic ratios of the two elements in simple binary compounds such as iron sulfide must have ratios of small whole numbers. The specified compound contains about equal numbers of iron and sulfur and therefore has the formula FeS.
What evidence indicates that a chemical change took place when the iron and sulfur combined to from iron sulfur?
Evidence of a chemical change occurring when iron and sulfur combine to form iron sulfide includes the observation of color change from gray/brownish to black, the release of heat energy as the reaction occurs, and the formation of a new substance with different properties than the original iron and sulfur. Additionally, the mass of the iron sulfide formed would be equal to the combined masses of the iron and sulfur used in the reaction, as mass is conserved in chemical reactions.
What happens to the mass in the reaction between iron and sulfur?
Total mass before reaction=Total mass after reaction Proven by the law of mass conversion: Matter cannot be created or destroyed; it can only change forms.
What mass of iron is needed to react with 16 grams of sulfur?
iron forms iron sulphide when reacts with sulphur, Fe + S = FeS in this reaction 32 g of sulphur require 56 g of iron so 16 g sulphur requires 28 g of iron.
In the reaction 2 Fe plus 3 S--- Fe2S3 how much iron is required to react with 48.0 g of sulfur?
To determine the amount of iron required, you first need to find the molar mass of sulfur (32.06 g/mol) and the molar mass of iron (55.85 g/mol). Then, use the stoichiometry of the reaction to calculate the amount of iron needed, which is 64.4 g.
What happens to the number of particles at the beginning and end of the iron and sulfur reaction?
In a reaction between iron and sulfur to form iron sulfide, the total number of particles remains the same before and after the reaction. This is due to the law of conservation of mass, which states that matter cannot be created or destroyed in a chemical reaction.
Why are there the same number of of particles at the beginning as at the end of the iron and sulfur reaction?
In a chemical reaction like iron reacting with sulfur to form iron sulfide, the number of atoms of each element in the reactants must equal the number of atoms of each element in the products. This is known as the principle of conservation of mass, which states that matter cannot be created or destroyed in a chemical reaction.
What is the mass of iron sulphide produced when 5.6g of iron completely reacts with excess sulfur?
The balanced chemical equation for the reaction between iron and sulfur to form iron sulfide is: 8 Fe + S8 -> 8 FeS Given that 5.6g of iron completely reacts with excess sulfur, we can calculate the mass of iron sulfide produced. Using stoichiometry, we find that 5.6g of Fe corresponds to 5.6g of FeS being produced, as the molar ratio between Fe and FeS is 1:1.
Which halogen has a greater mass then sulfur but lower mass than iron?
Sulfur atomic mass: 32 a.m.e.Chlorine atomic mass: 35.5 a.m.e.Iron atomic mass: 56 a.m.e.
When a mixture of iron and sulfur is heated a compound called iron sulfide is madein an experiment 2.8g of iron made 4.4g of iron sulfide What mass of sulfur reacted with the 2.8g of iron?
3.2g ..... assuming it is iron (II) Sulphide FeS2. The relative atomic mass of iron is 56, sulphur is 32, so the ratio of masses of these two elements IN THE FORMULA = 56 : 64. If 2.8g of iron was used then, from the ratio, this will react with 3.2g of sulphur.
How much iron sulphide will be formed if 56g of iron and 40g of sulphur are mixed?
Iron(II) sulfide is formed by the reaction of iron and sulfur as follows: Fe + S --> FeS To determine the amount of iron(II) sulfide formed, we need to consider the limiting reactant, which is sulfur in this case. The molar mass of iron(II) sulfide is 87.91 g/mol. Therefore, if 40g of sulfur reacts, it will form 40g / (32.06g/mol) = 1.25 moles of iron(II) sulfide, which is equivalent to 1.25 x 87.91 = 109.89g. So, 109.89g of iron sulfide will be formed.
What is the percent of sulfur in iron III sulfate?
It would depend on your definition of percentage. For example, if you were looking for percentage of atom count, you would count the atoms in the compound Fe2(SO4)3. You have 12 Oxygen, 3 Sulfur, and 2 Iron, leaving you with 3 sulfur out of 17 total atoms, or 3/17 or 0.176470588 (about 17.5%). If, on the other hand, you are talking about percentage of mass, then you must calculate the mass of each element in the compound. Iron's atomic mass is 55.847. Sulfur's atomic mass is 32.066. Oxygen's atomic mass is 15.999. Multiply these by their respective ratios. Iron (55.847*2=111.694) Sulfur (32.066*3=96.198) Oxygen (15.999*12=191.988) Add them together to get (111.694+96.198+191.988=399.88) Now, take the overall mass of the sulfur part (96.198) and divide it into the total mass (399.88). The answer is 96.198/399.88 or 0.24056717 or about 24%.
What type of reaction occurs and how does the mass of the object after rusting compare with its original mass?
Rusting is a chemical reaction known as oxidation. When a metal object rusts, it gains mass because the iron in the metal reacts with oxygen to form iron oxide (rust). The mass of the object after rusting will be greater than its original mass due to the addition of the iron oxide.
