BaO (s) + H2O (g) = Ba(OH)2 (s) If you want to completely react 10.25 g of barium oxide, how many grams of water will you need?

Molar Mass of barium hydroxide = 171.34g/mol5% of barium hydroxide = 8.567 g/molSulfur IV oxide Molar Mass = 64.066g/mol2 moles of sulfur IV oxide = 128.066g/molX×(1mol/128.066) × (2/1) = 2moles...

Since the mole ratio is equal 1:1 then the theoretical mass of Ba(OH)2 produced would be =2.34moles.

Molar mass of Ca(OH)2= 74.093 g/mol86.4/74.093= 1.166molMolar mass of HCl= 36.458 g/mol89.7/36.458= 2.46molMole ratio HCl: H2O= 2:1Mole ratio H2O:Ca(OH)2= 1:11.166+1.23=1.396molMolar mass of H2O = 18.0151.396×18.015= 25.15g

Grams of aluminum oxide produced =12.57g

9.97g of Barium

n(NO2) = 138/46 = 3 moln(H2O) = 54/18 = 3 molwater in excess, therefore:n(HNO3) = n(NO2)*2/3 = 2 molm(HNO3) = 2*63 = 126 g

x y 0.100 3Ba(s) + 2N2(g) ------gt; Ba3N2(s) 3 2 1 x=3*0.100/1=0.300 mol y=2*0.100/2-0.200 mol

Solution:Solid barium reacts with oxygen gas at a temperature of over 800°C to produce solid barium oxide.2Ba + O2 = 2BaO.

The chemical equation is 8Ba(s) + S8 (s) → 8BaS (s) Molar mass of Barium is M = 137.3 g/mol Molar mass of barium sulfide is M = 169.4 g/mol...

K2SO4 + BaCl2*2H2O = BaSO4 + 2 KCl + 2 H2O n (K2SO4) = m/M = (6.77 g)/(174 g/mol) = 0.039 mol n (BaCl2*2H2O) = n (K2SO4) = 0.039 mol...