The Atomic Mass of magnesium (Mg) is approximately 24.305 grams per mole.
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 number of moles in 13.5 grams of magnesium nitrate, you need to divide the given mass by the molar mass of magnesium nitrate. The molar mass of magnesium nitrate (Mg(NO3)2) is 148.31 g/mol. Moles of magnesium nitrate = 13.5 grams / 148.31 g/mol ≈ 0.091 moles
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.
The molar mass of magnesium is 24.3 g/mol and oxygen is 16 g/mol. The balanced equation is 2Mg + O2 -> 2MgO. Since magnesium and oxygen react in a 1:1 ratio, all magnesium reacted to form magnesium oxide. Therefore, 2.5 grams of magnesium oxide was formed.
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.
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.
Theoretically the mass is 62,3018 g.
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.
1,11 moles of magnesium have 26,97855 g.
There are 0.13 moles in 20 grams of magnesium nitrate.
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 ===================
88,1 moles of magnesium is equivalent to 2 141,27 g.
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To find the number of moles in 13.5 grams of magnesium nitrate, you need to divide the given mass by the molar mass of magnesium nitrate. The molar mass of magnesium nitrate (Mg(NO3)2) is 148.31 g/mol. Moles of magnesium nitrate = 13.5 grams / 148.31 g/mol ≈ 0.091 moles
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.
Mass in grams = no of moles x molecular mass. So, mass in grams = 5.2x 56 = 291.2g