I’m not sure but it has to be hydrogen
A. 1,5,3,4
the 3 and 4 increase the number of oxygens on the right side to 10 so you have to put a 5 in front of the oxygen of the left side to balance it out
There are at least 7 strong acids
Answer:
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
