1,4 moles of lead(II) oxide are formed.
There are 0.13 moles in 20 grams of magnesium nitrate.
The molecular weight of potassium nitrate (KNO3) is 101.1 grams/mole.K = 39.1 + N = 14.0 + O = 3*16.0 = 101.1Now divide 6.5 by 101.1 to calculate # of moles => 0.064 moles KNO3
These reagents doesn't react.
6 moles
20,32 g of lithium nitride can be obtained.
There are 0.13 moles in 20 grams of magnesium nitrate.
How many MOLES of sodium nitrate are present in 2.85 grams of this compound ?
how many moles are there in 56.0 grams of silver nitrate?
1,25 g of anhydrous iron(III) nitrate = 0,005 moles
15.451
Ca(NO3)2 5.600 grams calcium nitrate ( 1mole Ca(NO3)2/164.1 grams)(2 moles NO3/1 mole Ca(NO3)2) = 6.825 X 10 -2 moles of nitrate ions -----------------------------------------------
Hydrogen nitrate has a mass of 63.01 g/mol. In order to find the number of moles you divide the grams by the molar mass. 250/63.01 = 3.96 mol.
You have2KClO3 ==> 2KCl + 3O2 as the balanced equation 25 g KClO3 x 1 mole/123 g = 0.20 moles moles KCl formed = 0.20 moles KClO3 x 2 moles KCl/2 moles KClO3 = 0.20 moles KCl formed grams KCl = 0.20 moles x 74.5 g/mole = 14.9 g = 15 grams of KCl formed
4.50 g(1/169.88) =.0265 mols
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.
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.
The molecular weight of potassium nitrate (KNO3) is 101.1 grams/mole.K = 39.1 + N = 14.0 + O = 3*16.0 = 101.1Now divide 6.5 by 101.1 to calculate # of moles => 0.064 moles KNO3