6NaBr+1AlO3=3Na2O+2AlBr3 How many grams of NaBr would be needed in order to make 23.5 grams of AlBr3

23.5 grams of AlBr3 will be produced by 27.20 grams of NaBr

Explanation:

The balanced equation here is

6NaBr + 1AlO3 = 3Na2O + 2AlBr3

6 moles of NaBr are required to produce 2 moles of AlBr3

Mass of one mole of NaBr = 102.894 g/mol

Mass of one mole of AlBr3 = 266.69 g/mol

Mass of 6 moles of NaBr = 6*102.894 g/mol

Mass of two moles of AlBr3 = 2*266.69 g/mol

6*102.894 g  NaBr produces 2*266.69 g of AlBr3

23.5 grams of AlBr3 will be produced by (6*102.894)/(2*266.69 )*23.5 = 27.20 grams of NaBr

a chemical reaction happens

explanation:

number of molecules / avogadro's number * molecule molecular = mass

explanation:

we define a gram-molecular weight as the mass of one mole of any element or compound. a mole contains avogadro's number of units (molecules) –  

6.022

x

10

23

. so, dividing the number of molecules by avogadro's number gives us the number of moles. multiplying that by the molecular weight gives us the mass. from the periodic table we find the gram-molecular weight of u to be238 and of f to be19. so,  

u

f

6

will have a gram-molecular weight of 352g.