46 grams of sodium is 2 moles. 2 mol of sodium forms 1 mol of sodium oxide. So it makes 62 g of sodium oxide.
If the reaction is:6 Na + 2 O2 = 2 Na2O + Na2O2This mass is 3,83 g sodium.
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 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.
There are two elements, namely sodium and oxygen.
There are 2.54 grams of sodium in 1 gram of sodium carbonate.
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.
If the reaction is:6 Na + 2 O2 = 2 Na2O + Na2O2This mass is 3,83 g sodium.
62 grams a+
155.2 g
Since the balanced equation is 4Na + O2 -> 2NaO, the molar ratio between Na and NaO is 4:2 or 2:1. This means that for every 2 moles of Na used, 1 mole of NaO is formed. The molar mass of NaO is 62 g/mol. Therefore, if 46 grams of sodium are used, then 31 grams of sodium oxide will be formed.
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 amount of oxygen used, we need to consider the difference in mass between sodium and sodium oxide. The mass increase is 16g (62g - 46g) which corresponds to the amount of oxygen used from the air. Therefore, 16g of oxygen from the air were used.
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.
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 --------------------------------------------
There are two elements, namely sodium and oxygen.
There are 2.54 grams of sodium in 1 gram of sodium carbonate.
The complete question: Lead (II) oxide reacts with ammonia forming solid lead nitrogen gas and liquid water. 1.)How many grams of ammonia are consumed in the reaction of 75.0g lead (II) oxide? 2.) If 56.4g of lead are produced how many grams of nitrogen are also formed?