Balanced equation.
4Na + O2 ->2Na2O
14.6 grams Na (1 mole Na/22.99 grams)(1 mole O2/4 mole Na)(32.0 grams/1 mole O2)
= 5.08 grams oxygen gas needed
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62 grams a+
62 grams a+
5.0 grams.
66g
how much sodium hydroxide in grams must be added to seawater to precipitate 86.9mg of magnesium present?
The answer is 152 g oxygen.
999 g
5g of oxygen are required to completely react with 10g of hydrogen resulting in water with no net losses. Or exactly 2:1 = H20 (2 hydrogen < > 1 oxygen)
The equation for the reaction is 4 Na + O2 -> 2 Na2O. This shows that, for complete reaction, one mole of oxygen is required for each four gram atomic masses of sodium. The gram atomic mass of sodium is 22.9898; therefore, 46 grams of sodium constitutes 2.00 moles of sodium, to more than the justified number of significant digits. The gram molecular mass of diatomic oxygen is 31.9988; therefore 160 grams of oxygen constitutes 5.000 moles of diatomic oxygen, to more than the justified number of significant digits. This is well over the minimum amount of oxygen required for complete reaction of all the sodium present. Each two gram atomic masses of sodium produces one gram formula mass of sodium oxide; therefore, the number of gram formula masses of sodium oxide produced is 1.00, to at least the justified number of significant digits.
62 grams a+
62 grams a+
Infinite because Sodium Chloride is neutral and will not neutralize sulfuric acid.
800 g oxygen are needed.
Take the balanced equation CH4+2O2---->CO2+2H2O.So according to it 40g of O2
how many grams of hydrogen are required to completely saturate 75 grams oleic acid which has one double bond ?
5.0 grams.
66g