How many grams of sodium carbonate are produced when 5.3 moles of sodium phosphate reacts with aluminum carbonate?
1 answer:
Answer: There is 842.54 grams of sodium carbonate are produced when 5.3 moles of sodium phosphate reacts with aluminum carbonate.
Explanation:
Chemical equation depicting reaction between sodium phosphate and aluminum carbonate is as follows.
As this equation contains same number of atoms on both reactant and product side. So, this equation is a balanced equation.
According to the equation, 2 moles of sodium phosphate is giving 3 moles of sodium carbonate.
Therefore, sodium carbonate formed by 5.3 moles of sodium phosphate is as follows.
As number of moles is the mass of substance divided by its molar mass. So, mass of sodium carbonate ( molar mass = 105.98 g/mol) is as follows.
Thus, we can conclude that there is 842.54 grams of sodium carbonate are produced when 5.3 moles of sodium phosphate reacts with aluminum carbonate.
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Answer:
0.2g
Explanation:
Given parameters:
Number of moles of H₂ = 0.4mol
Number of moles of O₂ = 0.15mol
Unknown:
Mass of reactant that would remain = ?
Solution:
To solve this problem, we need to know the limiting reactant which is the one in short supply in the given reaction.
The expression of the reaction is :
2H₂ + O₂ → 2H₂O
2 mole of H₂ will combine with 1 mole of O₂
But given; 0.4 mole of H₂ we will require = 0.2mole of O₂
The given number of oxygen gas is 0.15mole and it is the limiting reactant.
Hydrogen gas is in excess;
1 mole of oxygen gas will combine with 2 mole of hydrogen gas
0.15 mole of oxygen gas will require 0.15 x 2 = 0.3mole of hydrogen gas
Now, the excess mole of hydrogen gas = 0.4 mole - 0.3 mole = 0.1mole
Mass of hydrogen gas = number of mole x molar mass
Molar mass of hydrogen gas = 2(1) = 2g/mol
Mass of hydrogen gas = 0.1 x 2 = 0.2g