Answer: Moles of hydrogen required are 4.57 moles to make 146.6 grams of methane, 
.
Explanation:
Given: Mass of methane = 146.6 g
As moles is the mass of a substance divided by its molar mass. So, moles of methane (molar mass = 16.04 g/mol) are calculated as follows.
The given reaction equation is as follows.
This shows that 2 moles of hydrogen gives 1 mole of methane. Hence, moles of hydrogen required to form 9.14 moles of methane is as follows.
Thus, we can conclude that moles of hydrogen required are 4.57 moles to make 146.6 grams of methane, .
Answer:
22.77 g.
he limiting reactant is O₂, and the excess reactant is Mg.
Explanation:
- From the balanced reaction:
<em>Mg + 1/2O₂ → MgO,</em>
1.0 mole of Mg reacts with 0.5 mole of oxygen to produce 1.0 mole of MgO.
- We need to calculate the no. of moles of (16.3 g) of Mg and (4.52 g) of oxygen:
no. of moles of Mg = mass/molar mass = (16.3 g)/(24.3 g/mol) = 0.6708 mol.
no. of moles of O₂ = mass/molar mass = (4.52 g)/(16.0 g/mol) = 0.2825 mol.
So. 0.565 mol of Mg reacts completely with (0.2825 mol) of O₂.
<em>∴ The limiting reactant is O₂, and the excess reactant is Mg (0.6708 - 0.565 = 0.1058 mol).</em>
<u><em>Using cross multiplication:</em></u>
1.0 mole of Mg produce → 1.0 mol of MgO.
∴ 0.565 mol of Mg produce → <em>0.565 mol of MgO.</em>
<em>∴ The amount of MgO produced = no. of moles x molar mass </em>= (0.565 mol)(40.3 g/mol) = <em>22.77 g.</em>
X
H has a positive 1 charge. This means that having 3H = +3<span>. This is a neutral compound so x= -3 because X+3H= 0
Y</span>
is also neutral so 2X+Y= 0
we know X=-3 So, 2(-3)+Y=0
-6+y=0
Y=+6 charge
Answer: The valency of X is -3.
The valency of Y is 6
Answer:
20 L of Cl₂
Solution:
The reaction is as follow,
H₂C₂ + 2 Cl₂ → H₂C₂Cl₄
According to equation,
167.84 g (1 mole) H₂C₂Cl₄ is produced by = 44.8 L (2 mole) of Cl₂
So,
75 g of H₂C₂Cl₄ will be produced by = X L of Cl₂
Solving for X,
X = (44.8 L × 75 g) ÷ 167.84 g
X = 20 L of Cl₂
Al 1s²2s²2p⁶3s²3p¹ - 3e⁻ → Al³⁺ 1s²2s²2p⁶
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