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:
357 g of the transition metal are present in 630 grams of the compound of the transition metal and iodine
Explanation:
In any sample of the compound, the percentage by mass of the transition metal is 56.7%. This means that for a 100 g sample of the compound, 56.7 g is the metal while the remaining mass, 43.3 g is iodine.
Given mass of sample compound = 630 g
Calculating the mass of iodine present involves multiplying the percentage by mass composition of the metal by the mass of the given sample;
56.7 % = 56.7/100 = 0.567
Mass of transition metal = 0.567 * 630 = 357.21 g
Therefore, the mass of the transition metal present in 630 g of the compound is approximately 357 g
The first step to answering this item is to convert the given temperatures in °F to °C through the equation,
°C = (°F - 32)(5/9)
initial temperature: 72°F
°C = (72 - 32)(5/9) = 22.22°C
final temperature: 145°F
°C = (145 - 32)(5/9) = 62.78°C
Substituting to the equation,
H = mcpdT
H = (43 g)(0.903 J/g°C)(62.78 - 22.22)
H = 1574.82 J
<em>Answer: 1574.82 J</em>
