Masses: Mg= 24, S= 32, O= 16
Magnesium = 24g
Magnesium sulphate = (24+32)+(16x4) = 120g.
Therefor we know: 24g of Mg ----> makes 120g of MgSO4.
To get to 4g from 24g, you devide by 6.
(24 devided by 6 = 4g)
And to find what it would make, you devide 120 by 6 too.
Which = 20g.
4g of Mg ----> 20g of MgSO4.
To calculate the number of moles of magnesium used, you divide the mass of magnesium by its molar mass. The molar mass of magnesium is approximately 24.31 g/mol. For example, if you have 12.15 grams of magnesium, you would divide 12.15 by 24.31 to find that you have 0.5 moles of magnesium.
To determine the amount of oxygen, we first find the amount of magnesium by subtracting the given 20.0 grams of magnesium oxide from the total. Given that the molar mass of magnesium oxide is 40.3 g/mol and that of magnesium is 24.3 g/mol, we calculate the amount of oxygen by adjusting accordingly. This process gives us the weight ratio of magnesium oxide to oxygen.
To find the mass of magnesium sulfate needed, first calculate the moles of MgSO4 required using the formula: moles = molarity × volume (in liters). Then, multiply the moles by the molar mass of MgSO4 (120.37 g/mol) to get the mass in grams. In this case, the mass of MgSO4 needed would be approximately 450.6 grams.
4.00 grams of magnesium oxide is composed of 2.43 grams of magnesium (Mg) and 1.57 grams of oxygen (O). Therefore, to produce 4.00 grams of magnesium oxide, you would need 2.43 grams of magnesium.
Approximately 73.2 grams of copper sulphate can be dissolved in 50 grams of water at 60 degrees Celsius. This is the maximum amount of copper sulphate that the water can hold in a saturated solution at that temperature.
To calculate the number of moles of magnesium used, you divide the mass of magnesium by its molar mass. The molar mass of magnesium is approximately 24.31 g/mol. For example, if you have 12.15 grams of magnesium, you would divide 12.15 by 24.31 to find that you have 0.5 moles of magnesium.
Balanced equation first. 3Mg + N2 -> Mg3N2 55.3 grams Mg (1 mole Mg/24.31 grams)(1 mole Mg3N2/3 mole Mg)(100.95 grams/1 mole Mg3N2) = 76.5 grams Mg3N2 made ===================
To determine the amount of oxygen, we first find the amount of magnesium by subtracting the given 20.0 grams of magnesium oxide from the total. Given that the molar mass of magnesium oxide is 40.3 g/mol and that of magnesium is 24.3 g/mol, we calculate the amount of oxygen by adjusting accordingly. This process gives us the weight ratio of magnesium oxide to oxygen.
400 grams of nickel sulphate (anhydrous) is equivalent to 2,58 moles.
4.00 grams of magnesium oxide is composed of 2.43 grams of magnesium (Mg) and 1.57 grams of oxygen (O). Therefore, to produce 4.00 grams of magnesium oxide, you would need 2.43 grams of magnesium.
To find the mass of magnesium sulfate needed, first calculate the moles of MgSO4 required using the formula: moles = molarity × volume (in liters). Then, multiply the moles by the molar mass of MgSO4 (120.37 g/mol) to get the mass in grams. In this case, the mass of MgSO4 needed would be approximately 450.6 grams.
2
If 3 grams of magnesium are used to form 4 grams of magnesium oxide, then 1 gram of oxygen is used in the reaction. This means 1 gram of oxygen remains unused.
The atomic mass of magnesium (Mg) is approximately 24.305 grams per mole.
Approximately 73.2 grams of copper sulphate can be dissolved in 50 grams of water at 60 degrees Celsius. This is the maximum amount of copper sulphate that the water can hold in a saturated solution at that temperature.
calculate the percent yiel if 49 grams of PCL5 is obtained from the treatment of 33.9 grams of pcl5 with 19 grams cl2
Theoretically the mass is 62,3018 g.