The volume of H₂ : = 15.2208 L
<h3>Further explanation</h3>
Given
Reaction
2 As (s) + 6 NaOH (aq) → 2 Na₃AsO₃ (s) + 3 H₂ (g)
34.0g of As
Required
The volume of H₂ at STP
Solution
mol As (Ar = 75 g/mol) :
= mass : Ar
= 34 g : 75 g/mol
= 0.453 mol
From the equation, mol ratio As : H₂ = 2 : 3, so mol H₂ :
=3/2 x mol As
=3/2 x 0.453
= 0.6795
At STP, 1 mol = 22.4 L, so :
= 0.6795 x 22.4 L
= 15.2208 L
The answer is C. and D. water and aluminum
Answer:
≅ 16.81 kJ
Explanation:
Given that;
mass of acetone = 31.5 g
molar mass of acetone = 58.08 g/mol
heat of vaporization for acetone = 31.0 kJ/molkJ/mol.
Number of moles = 
Number of moles of acetone = 
Number of moles of acetone = 0.5424 mole
The heat required to vaporize 31.5 g of acetone can be determined by multiplying the number of moles of acetone with the heat of vaporization of acetone;
Hence;
The heat required to vaporize 31.5 g of acetone = 0.5424 mole × 31.0 kJ/mol
The heat required to vaporize 31.5 g of acetone = 16.8144 kJ
≅ 16.81 kJ
Answer:
HCl
Explanation:
Given data:
Mass of Zn = 50 g
Mass of HCl = 50 g
Limiting reactant = ?
Solution:
Chemical equation:
Zn + 2HCl → ZnCl₂ + H₂
Number of moles of Zn:
Number of moles = mass / molar mass
Number of moles = 50 g/ 65.38 g/mol
Number of moles = 0.76 mol
Number of moles of HCl:
Number of moles = mass / molar mass
Number of moles = 50 g/ 36.5 g/mol
Number of moles = 1.4 mol
Now we will compare the moles of Reactant with product.
Zn : ZnCl₂
1 : 1
0.76 : 0.76
Zn : H₂
1 : 1
0.76 : 0.76
HCl : ZnCl₂
2 : 1
1.4 : 1/2×1.4 = 0.7
HCl : H₂
2 : 1
1.4 : 1/2×1.4 = 0.7
Less number of moles of product are formed by HCl it will act limiting reactant.
They can be differnt shape or one must weight more then the other one
