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To calculate the amount of iron in Fe2S3, first find the molar mass of Fe2S3: Iron (Fe) has a molar mass of 55.85 g/mol and sulfur (S) has a molar mass of 32.06 g/mol. So, the molar mass of Fe2S3 is 2(55.85) + 3(32.06) = 222.92 g/mol. Next, divide the molar mass of iron by the molar mass of Fe2S3 and multiply by the given mass of Fe2S3 to find the grams of iron: (2 * 55.85 g/mol / 222.92 g/mol) * 34 g = 8.53 g of iron in 34 g of Fe2S3.
To find the moles of Fe2S3 produced, convert 0.690 g to moles using the molar mass of Fe2S3. Then, use the stoichiometry of the reaction to determine the moles of FeCl3 required. Finally, use the molarity of FeCl3 to calculate the volume in milliliters needed, keeping in mind the percent yield.
To determine the grams of potassium chloride formed, you first need to calculate the moles of oxygen produced by the decomposition of potassium chlorate. Then, use the stoichiometry of the balanced chemical equation to convert moles of oxygen to moles of potassium chloride. Finally, from the molar mass of potassium chloride, you can calculate the grams formed.
To find the mass of BrCl formed, you first need to determine the limiting reactant by comparing the mole ratios of Cl2 and Br2 in the balanced equation. Once you know the limiting reactant, you can use stoichiometry to calculate the moles of BrCl formed. Finally, convert the moles of BrCl to grams using the molar mass of BrCl.
To find the grams of uranium oxide formed, we need to determine the molar mass of uranium and oxygen, calculate the moles of each element present, and finally the moles of uranium oxide formed. Then, we convert moles to grams using the molar mass of uranium oxide. The final answer for the grams of uranium oxide formed depends on the stoichiometry of the reaction.
To calculate the amount of iron in Fe2S3, first find the molar mass of Fe2S3: Iron (Fe) has a molar mass of 55.85 g/mol and sulfur (S) has a molar mass of 32.06 g/mol. So, the molar mass of Fe2S3 is 2(55.85) + 3(32.06) = 222.92 g/mol. Next, divide the molar mass of iron by the molar mass of Fe2S3 and multiply by the given mass of Fe2S3 to find the grams of iron: (2 * 55.85 g/mol / 222.92 g/mol) * 34 g = 8.53 g of iron in 34 g of Fe2S3.
.2M x V FeCl3=moles FeCl3 x 1mole Fe2S3/2mole FeCl3=moles of Fe2S3 x mm of Fe2S3/1 mole Fe2S3= g Fe2S3 x .65% yield. 2.75g Fe2S3/ .65= 4.23g Fe2S3/ 207.91= .02035 x 2mole FeCl3=.0407 moles FeCl3/ .2M FeCl3= .2035 L x 1000= 203.5 ml
To determine the amount of iron that can be produced from 119 g of Fe2S3 and 12.7 g of C, we first need to calculate the molar mass of Fe2S3 and C. The molar mass of Fe2S3 is approximately 207.9 g/mol, and the molar mass of C is approximately 12.01 g/mol. Next, we calculate the moles of Fe2S3 and C by dividing the given masses by their respective molar masses. Then, we determine the limiting reactant by comparing the moles of Fe2S3 and C. Finally, we use the stoichiometry of the balanced chemical equation to calculate the theoretical yield of iron, which is approximately 42.4 grams.
To find the moles of Fe2S3 produced, convert 0.690 g to moles using the molar mass of Fe2S3. Then, use the stoichiometry of the reaction to determine the moles of FeCl3 required. Finally, use the molarity of FeCl3 to calculate the volume in milliliters needed, keeping in mind the percent yield.
To determine the grams of potassium chloride formed, you first need to calculate the moles of oxygen produced by the decomposition of potassium chlorate. Then, use the stoichiometry of the balanced chemical equation to convert moles of oxygen to moles of potassium chloride. Finally, from the molar mass of potassium chloride, you can calculate the grams formed.
To determine the number of grams of water formed, we need to calculate the moles of butanol (C4H9OH) and then use the balanced chemical equation to find the moles of water produced in the combustion reaction. From there, we convert moles of water to grams. The balanced equation for the combustion of butanol is C4H9OH + 6O2 → 4CO2 + 5H2O.
The formula of iron (III) sulfide is Fe2S3, showing that each formula unit contains two iron atoms. The gram formula unit mass for iron (III) sulfide is 207.87, and the gram atomic mass of iron is 55.847. Therefore, the fraction by mass of iron in iron (III) sulfide is 2(55.847)/207.87 or about 0.5373, and the grams of iron in 130 g of iron (III) sulfide is 0.5373 X 130 or 69.9 grams, to the justified number of significant digits.
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530,3 g potassium iodide are needed.
To find the mass of BrCl formed, you first need to determine the limiting reactant by comparing the mole ratios of Cl2 and Br2 in the balanced equation. Once you know the limiting reactant, you can use stoichiometry to calculate the moles of BrCl formed. Finally, convert the moles of BrCl to grams using the molar mass of BrCl.
89,6 g ammonia are obtained.
To find the grams of uranium oxide formed, we need to determine the molar mass of uranium and oxygen, calculate the moles of each element present, and finally the moles of uranium oxide formed. Then, we convert moles to grams using the molar mass of uranium oxide. The final answer for the grams of uranium oxide formed depends on the stoichiometry of the reaction.