One mole of anhydrous Cu(NO3)2 has 187,5558 g.
To produce 1 mole of urea, 1 mole of carbon dioxide is needed. The molar mass of urea is 60 grams/mol, and the molar mass of carbon dioxide is 44 grams/mol. Therefore, to produce 125 grams of urea, 125 grams/60 grams/mol = 2.08 moles of urea is needed. This means 2.08 moles of carbon dioxide is needed, which is 2.08 moles * 44 grams/mol = 91.52 grams of carbon dioxide needed.
The balanced chemical equation for the reaction between iron and oxygen to produce Fe2O3 is 4Fe + 3O2 -> 2Fe2O3. From the equation, we see that 3 moles of oxygen react with 4 moles of iron to produce 2 moles of Fe2O3. Therefore, to find the grams of oxygen needed, we need to calculate the molar mass of Fe2O3 and then determine the number of grams needed using the mole ratio from the balanced equation.
To find the mass of oxygen needed to produce 13 grams of the oxide, we first find the molar ratio between P and O in the product. Then, use this ratio to calculate the mass of oxygen needed. Since 71g of the oxide contains 31g of phosphorus, we can calculate the mass of oxygen needed for 13g of the oxide by setting up a proportion.
To find the amount of aluminum needed to produce aluminum sulfate, you need to consider the molar mass of aluminum sulfate and the ratio of aluminum in the compound. First, calculate the molar mass of aluminum sulfate (Al2(SO4)3). Then, find the ratio of aluminum in the compound (2 moles of Al in 1 mole of Al2(SO4)3). Finally, use this information to calculate the grams of aluminum needed to produce 25.0 grams of aluminum sulfate.
To produce 1 mole of chloroform, you need 3 moles of chlorine. So, to produce 1.5 moles of chloroform, you would need 4.5 moles of chlorine. Converting moles to grams by using the molar mass of chlorine (35.5 g/mol) gives you 160.5 grams of chlorine required.
To produce 1 gram of DDT, 3.3 grams of chloral are needed. Therefore, to produce 10.5 grams of DDT, you would need 10.5 * 3.3 = 34.65 grams of chloral.
To produce 1 mole of urea, 1 mole of carbon dioxide is needed. The molar mass of urea is 60 grams/mol, and the molar mass of carbon dioxide is 44 grams/mol. Therefore, to produce 125 grams of urea, 125 grams/60 grams/mol = 2.08 moles of urea is needed. This means 2.08 moles of carbon dioxide is needed, which is 2.08 moles * 44 grams/mol = 91.52 grams of carbon dioxide needed.
The balanced chemical equation for the reaction between iron and oxygen to produce Fe2O3 is 4Fe + 3O2 -> 2Fe2O3. From the equation, we see that 3 moles of oxygen react with 4 moles of iron to produce 2 moles of Fe2O3. Therefore, to find the grams of oxygen needed, we need to calculate the molar mass of Fe2O3 and then determine the number of grams needed using the mole ratio from the balanced equation.
16,45 g nitrogen are needed.
A lot
800 g oxygen are needed.
530,3 g potassium iodide are needed.
To find the mass of oxygen needed to produce 13 grams of the oxide, we first find the molar ratio between P and O in the product. Then, use this ratio to calculate the mass of oxygen needed. Since 71g of the oxide contains 31g of phosphorus, we can calculate the mass of oxygen needed for 13g of the oxide by setting up a proportion.
No. Just the opposite. Every kilogram is a package of 1,000 grams. So naturally, if you pack all of your grams into packages of 1,000, your large number of grams will produce a smaller number of kilogram-packages.
To find out how many grams of N2 are needed to produce 1.7 grams of NH3, you need to look at the balanced chemical equation for the reaction. For the reaction N2 + 3H2 -> 2NH3, the molar ratio of N2 to NH3 is 1:2. So you would need half the number of grams of N2 as NH3, which is 0.85 grams of N2.
To find the grams of CCl4 needed, multiply the number of moles (5.000 mol) by the molar mass of CCl4 (153.82 g/mol). (5.000 , \text{mol} \times 153.82 , \text{g/mol} = 769.1 , \text{g}) Therefore, 769.1 grams of CCl4 are needed.
To find the amount of aluminum needed to produce aluminum sulfate, you need to consider the molar mass of aluminum sulfate and the ratio of aluminum in the compound. First, calculate the molar mass of aluminum sulfate (Al2(SO4)3). Then, find the ratio of aluminum in the compound (2 moles of Al in 1 mole of Al2(SO4)3). Finally, use this information to calculate the grams of aluminum needed to produce 25.0 grams of aluminum sulfate.