Answer:
Mass = 2.89 g
Explanation:
Given data:
Mass of NH₄Cl = 8.939 g
Mass of Ca(OH)₂ = 7.48 g
Mass of ammonia produced = ?
Solution:
2NH₄Cl + Ca(OH)₂ → CaCl₂ + 2NH₃ + 2H₂O
Number of moles of NH₄Cl:
Number of moles = mass/molar mass
Number of moles = 8.939 g / 53.5 g/mol
Number of moles = 0.17 mol
Number of moles of Ca(OH)₂ :
Number of moles = mass/molar mass
Number of moles = 7.48 g / 74.1 g/mol
Number of moles = 0.10 mol
Now we will compare the moles of ammonia with both reactant.
NH₄Cl : NH₃
2 : 2
0.17 : 0.17
Ca(OH)₂ : NH₃
1 : 2
0.10 : 2/1×0.10 = 0.2 mol
Less number of moles of ammonia are produced by ammonium chloride it will act as limiting reactant.
Mass of ammonia:
Mass = number of moles × molar mass
Mass = 0.17 mol × 17 g/mol
Mass = 2.89 g
Answer:
i belive A, i cold be wrong tho, pls tell me if im right if you fully find out
ght. Explanation:
Answer:
2 Fe(iii)2O3 + 3 C ==> 2 Fe + 3 CO2
Explanation:
First of all, you have to translate the words into an equation.
Fe(iii)2O3 + C ==> Fe + CO2
The easiest way to tackle this is to start with the Oxygens and balance them. They must balance by going to the greatest common factor which is 6. So you multiply the molecule by whatever it takes to get the Oxygens to 6
2 Fe(iii)2O3 + C ==> Fe + 3 CO2
Now work on the irons. There 2 on the left and just 1 on the right. So you need to multiply the iron by 2.
2 Fe(iii)2O3 + C ==> 2 Fe + 3 CO2
Finally it is the turn of the carbons. There are 3 on the right, so you must make the carbon on the left = 3
2 Fe(iii)2O3 + 3 C ==> 2 Fe + 3 CO2
And you are done.
