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2.4088 x 1024 nitrate ions
Mg2+(s) + 2HNO3(l)= Mg(NO3)2(aq) + H2(g) since the only mole value given is 8 I must assume this is the limiting reactant. Because of the 2:1 ratio of Nitric acid to Magnesium Nitrate, meaning there must be 2 moles Nitric acid for every 1 mole Magnesium Nitrate formed, 4 moles of Magnesium nitrate will be formed.
I assume that is 28.0 grams.28.0 grams MgCl2 (1 mole MgCl2/95.21 grams)= 0.294 moles magnesium chloride==========================Now,One mole Mg 2+ = 0.294 moles Mg 2+ ions=============================2 moles Cl - = 0.588 moles Cl - ions=========================
13.5g Mg(NO3)2 x 1 mol Mg(NO3)2/148.3 = 0.0910 mol Mg(NO3)2
A sample of 56,4 of magnesium is equal to 2,32 moles.
There are 0.13 moles in 20 grams of magnesium nitrate.
2.4088 x 1024 nitrate ions
2.4088 x 1024 nitrate ions
Mg2+(s) + 2HNO3(l)= Mg(NO3)2(aq) + H2(g) since the only mole value given is 8 I must assume this is the limiting reactant. Because of the 2:1 ratio of Nitric acid to Magnesium Nitrate, meaning there must be 2 moles Nitric acid for every 1 mole Magnesium Nitrate formed, 4 moles of Magnesium nitrate will be formed.
The formula for magnesium oxide is MgO, showing that each formula unit of magnesium oxide contains one mole of magnesium ions. Therefore, if there is ample oxygen available, 4 moles of magnesium will form 4 moles of magnesium oxide.
I assume that is 28.0 grams.28.0 grams MgCl2 (1 mole MgCl2/95.21 grams)= 0.294 moles magnesium chloride==========================Now,One mole Mg 2+ = 0.294 moles Mg 2+ ions=============================2 moles Cl - = 0.588 moles Cl - ions=========================
13.5g Mg(NO3)2 x 1 mol Mg(NO3)2/148.3 = 0.0910 mol Mg(NO3)2
Two Chloride ions (2Cl-) ions are needed with their -1 charge on each one to cancel out the +2 charge of the single Magnesium ion (Mg2+). So Magnesium Chloride would have the chemical formula: MgCl2
there are two moles produced in potassium nitrate.
A sample of 56,4 of magnesium is equal to 2,32 moles.
Calcium Nitrtae is Ca(NO3)2 and so there are two moles of nitrate per mole of calcium nitrate. Thus there are 2 x 2.50 = 5.0 moles of nitrate present.
0.0999458293608 moles