The mass of heavy water produced when 7,00grams of oxygen reacts with excess D2 is 7,875 g.
The equation is 2C4H10 + 13O2 --> 8CO2 + 10H2O This means that for each mole of butane there are 5 moles of water produced. We have 7.01 g of butane = 7.01/58 moles of butane = 0.12 moles. Thus we will get 5 x 0.12 moles of water, = 5 x 0.12 x 18 g of water = 10.88 g.
When barium reacts with oxygen barium oxide is produced. Ba + O2 -> BaO
Rust.
0.95 - 0.954
nitrogen oxide
calcium oxide, CaO
oxygen
gay
2Na(s)+O2(g) ----> Na2O2(s)
If everything is under gaseous condition and with equal pressure and temperature, then also 4.0 L SO3 gas is produced from 4.0 L SO2,g.
Nothing, the reactants and products all are colourless.
Methane + Oxygen > Carbon (soot) + Water