30.8g of iron (approximately) reacted with 13.2g of oxygen will yield 44g of iron(III) oxide (Fe2O3) with 2.8g of oxygen left unreacted. This assumes Atomic Mass numbers of 56 and 16 respectively for iron and oxygen. The actual mass number of iron is 55.847 and oxygen 15.9994 making the figures 30.775g of iron and 13.225g of oxygen with 2.775g of unreacted oxygen. Of course this is an exothermic reaction so will there be a tiny tiny loss of mass in the system as it is converted to heat energy, according to E=MC^2? I'll let you work that one out...
To find the number of moles in 28 grams of calcium oxide, we need to divide the given mass by the molar mass of calcium oxide. The molar mass of calcium oxide (CaO) is 56.08 g/mol. So, 28 grams of CaO is equal to 28 g / 56.08 g/mol = 0.5 moles of calcium oxide.
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 --------------------------------------------
When aluminum, nitrogen, and oxygen are combined, the compound formed is aluminum nitride oxide (AlNO).
An oxygen atom can only be oxidized by some element with an equal or higher electronegativity. So, oxygen can be oxidized by fluorine (e.g., OF2) or by another oxygen atom (e.g., O2).
The balanced equation for the reaction is: 4 Na + O2 -> 2 Na2O. From the equation, 4 moles of sodium will react to form 2 moles of sodium oxide. Calculate the molar mass of Na2O (sodium oxide) to find out how many grams will be formed.
Sounds like 42.2 grams, but I haven't checked your proportions, are these stoichometric?
2
To find the grams of uranium oxide formed, we need to determine the molar mass of uranium and oxygen, calculate the moles of each element present, and finally the moles of uranium oxide formed. Then, we convert moles to grams using the molar mass of uranium oxide. The final answer for the grams of uranium oxide formed depends on the stoichiometry of the reaction.
To find the mass of oxygen needed to produce 13 grams of the oxide, we first find the molar ratio between P and O in the product. Then, use this ratio to calculate the mass of oxygen needed. Since 71g of the oxide contains 31g of phosphorus, we can calculate the mass of oxygen needed for 13g of the oxide by setting up a proportion.
If 3 grams of magnesium are used to form 4 grams of magnesium oxide, then 1 gram of oxygen is used in the reaction. This means 1 gram of oxygen remains unused.
To determine the amount of oxygen, we first find the amount of magnesium by subtracting the given 20.0 grams of magnesium oxide from the total. Given that the molar mass of magnesium oxide is 40.3 g/mol and that of magnesium is 24.3 g/mol, we calculate the amount of oxygen by adjusting accordingly. This process gives us the weight ratio of magnesium oxide to oxygen.
Sulphur + oxygen = Sulphur Oxide
The radium oxide is formed.
If 20g of mercury oxide were heated, the combined mass of oxygen and mercury would be 20 grams.
Iron oxide is formed by the reaction of iron and oxygen in a 1:1 ratio by mass. Therefore, the 55 g of iron will react completely with 55 g of oxygen to form iron oxide.
The molar mass of magnesium is 24.3 g/mol and oxygen is 16 g/mol. The balanced equation is 2Mg + O2 -> 2MgO. Since magnesium and oxygen react in a 1:1 ratio, all magnesium reacted to form magnesium oxide. Therefore, 2.5 grams of magnesium oxide was formed.
To determine the grams of aluminum oxide formed, we need to consider the balanced chemical equation for the reaction between aluminum and oxygen. The molar ratio between aluminum and aluminum oxide is 4:2. So, first calculate the moles of aluminum in 1020g, then use this to find the moles of aluminum oxide produced, and finally convert moles of aluminum oxide to grams.