Answer:
34.15% is the mass percentage of calcium in the limestone.
Explanation:
Mass of precipitate that is calcium oxalate = 140.2 mg = 0.1402 g
1 mg = 0.001 g
Moles of calcium oxalate = 
1 mole of calcium oxalate have 1 mole of calcium atom.
Then 0.001095 moles of calcium oxalate will have 0.001095 moles of calcium atom.
Mass of 0.001095 moles of calcium :
0.001095 mol × 40 g/mol = 0.04381 g
Mass of sample of limestone = 128.3 mg = 0.1283 g
Percentage of calcium in limestone:
34.15% is the mass percentage of calcium in the limestone.
Answer:
Explanation:
2 HCl(g) + Mg(s) → MgCl₂(s) + H₂(g)
Let's calculate the quantity of mole of produced hydrogen with the Ideal Gases Law
P . V = n . R .T
2.19 atm . 6.82L = n . 0.082 . 308K
(2.19 atm . 6.82L) / (0.082 . 308K) = n
0.591 mol = n
1 mol of H₂ gas came from 2 mol of hydrochloric, so, 0.591 mol came from the double of mole
0.591 .2 = 1.182 mole of acid.
Molar mass of HCl = 36.45 g/m
1.182 mole are (36.45 g/m . 1.182g ) contained in 43.1 g
Density HCl = HCl mass / HCl volume
0,118 g/mL = 43.1 g / HCl volume
43.1 g / 0.118 g/mL = 365.3 mL (HCl volume)
7. An exothermic reaction
8. The bonds are forming
Answer:
10.87 g of Ethyl Butyrate
Solution:
The Balance Chemical Equation is as follow,
H₃C-CH₂-CH₂-COOH + H₃C-CH₂-OH → H₃C-CH₂-CH₂-COO-CH₂-CH₃ + H₂O
According to equation,
88.11 g (1 mol) Butanoic Acid produces = 116.16 g (1 mol) Ethyl Butyrate
So,
8.25 g Butanoic Acid will produce = X g of Ethyl Butyrate
Solving for X,
X = (8.25 g × 116.16 g) ÷ 88.11 g
X = 10.87 g of Ethyl Butyrate
Answer:
Mass = 381.28 g
Explanation:
Given data:
Number of moles of HNO₃ = 16 mol
Mass of Cu needed to react with 16 mol of HNO₃ = ?
Solution:
Chemical equation:
3Cu + 8HNO₃ → 3Cu(NO₃)₂ + 4H₂O + 2NO
Now we will compare the moles of Cu with HNO₃ from balance chemical equation.
HNO₃ : Cu
8 : 3
16 : 3/8×16 = 6
Mass of Cu needed:
Mass = number of moles × molar mass
Mass = 6 mol × 63.546 g/mol
Mass = 381.28 g
