One would be phosporous whose configuration is 1s2 2s2 2p6 3s2 3p3
<em>m Na₂CO₃: 23g×2 + 12g + 16g×3 = 106 g/mol</em>
------------------------------
1 mol ------- 106g
X ------------ 10,6g
X = 10,6/106
<u>X = 0,1 mol Na₂CO₃</u>
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Explanation:
Answer:
"2.48 mole" of H₂ are formed. A further explanation is provided below.
Explanation:
The given values are:
Mole of Al,
= 3.22 mole
Mole of HBr,
= 4.96 mole
Now,
(a)
The number of mole of H₂ are:
⇒ 
or,
⇒ 
⇒ 
⇒ 
(b)
The limiting reactant is:
= 
(c)
The excess reactant is:
= 
Answer:
3.09kg
Explanation:
First, let us write a balanced equation for the reaction. This is illustrated below:
2C8H18 + 25O2 —> 16CO2 + 18H2O
Molar Mass of C8H18 = (12x8) + (18x1) = 96 + 18 = 114g/mol
Mass of C8H18 from the balanced equation = 2 x 114 = 228g
Converting 228g of C8H18 to kg, we obtained:
228/1000 = 0.228kg
Molar Mass of CO2 = 12 + (2x16) = 12 + 32 = 44g/mol
Mass of CO2 from the balanced equation = 16 x 44 = 704g
Converting 704g of CO2 to kg, we obtained:
704/1000 = 0.704kg
From the equation,
0.228kg of C8H18 produced 0.704kg of CO2.
Therefore, 1kg of C8H18 will produce = 0.704/0.228 = 3.09kg of CO2
