mass of pentane : = 30.303 g
moles of Al₂(CO₃)₃ : = 0.147
<h3>Further explanation</h3>
Given
1. Reaction
C₅H₁₂+8O₂→6H₂O+5CO₂.
45.3 g water
2. 2AlCl₃ + 3MgCO₃ → Al₂(CO₃)₃ + 3MgCl₂
37.2 MgCO₃
Required
mass of pentane
moles of Al₂(CO₃)₃
Solution
1. mol water = 45.3 : 18 g/mol = 2.52
From equation, mol ratio of C₅H₁₂ : H₂O = 1 : 6, so mol pentane :
= 1/6 x mol H₂O
= 1/6 x 2.52
= 0.42
Mass pentane :
= mol x MW
= 0.42 x 72.15 g/mol
= 30.303 g
2. mol MgCO₃ : 37.2 : 84,3139 g/mol = 0.44
mol Al₂(CO₃)₃ :
= 1/3 x mol MgCO₃
= 1/3 x 0.44
= 0.147
Answer:
0.1313 g.
Explanation:
- It is known that at STP, 1.0 mole of ideal gas occupies 22.4 L.
- Suppose that hydrogen behaves ideally and at STP conditions.
<u><em>Using cross multiplication:</em></u>
1.0 mol of hydrogen occupies → 22.4 L.
??? mol of hydrogen occupies → 1.47 L.
∴ The no. of moles of hydrogen that occupies 1.47 L = (1.0 mol)(1.47 L)/(22.4 L) = 6.563 x 10⁻² mol.
- Now, we can get the no. of grams of hydrogen in 6.563 x 10⁻² mol:
<em>The no. of grams of hydrogen = no. of hydrogen moles x molar mass of hydrogen</em> = (6.563 x 10⁻² mol)(2.0 g/mol) = <em>0.1313 g.</em>
Answer:
the utencils of a plant is called photosynthesis and your wc
Answer:
Explanation:
You would have to add up the atomic masses of all the compounds in the compound, making sure you include how many molecules of each are in the compound
For example, in CuSOA we have 1 molecule of Cu and S, as 4 molecules of O
The atomic masses are as follows:
Cu = 63.55 u
S = 32.065 u
O = 15.99 units
This is how we would add it up:
(Atomic mass of Cu) + (Atomic mass of S) + 4(Atomic Mass of O)
(63.55) + (32.065) + 4(15.99)
(63.55) + (32.065) + 63.96
= 159.575 u
