Answer:
A. 4.5 mol Mg(OH)₂
B. 6 mol NaOH
Explanation:
Let's consider the following balanced equation.
Mg(NO₃)₂ + 2 NaOH ⇒ Mg(OH)₂ + 2 NaNO₃
PART A
The molar ratio of NaOH to Mg(OH)₂ is 2:1. The moles of Mg(OH)₂ produced from 9 moles of NaOH are:
9 mol NaOH × 1 mol Mg(OH)₂/2 mol NaOH = 4.5 mol Mg(OH)₂
PART B
The molar ratio of NaOH to NaNO₃ is 2:2. The moles of NaOH needed to produce 6 moles of NaNO₃ are:
6 mol NaNO₃ × 2 mol NaOH/2 mol NaNO₃ = 6 mol NaOH
Answer:
S= 2(1) = 2
O= 2(4) = 8
Na= 2(2) = 4
Explanation:
The given compound is:
2Na₂SO₄
An element is a distinct substance that cannot be split up into simpler substances.
So;
Number of atoms of elements here are:
S= 2(1) = 2
O= 2(4) = 8
Na= 2(2) = 4
2.998e^8 is how I would write it. If you want it with the least amount of decimals, use 3e^8
Answer : The enthalpy of the reaction = -1839.6 KJ
Solution : Given,
= -520.0 KJ/mole
= -1699.8 KJ/mole
The balanced chemical reaction is,

Formula used :


We know that the standard enthalpy of formation of the element is equal to Zero.
Therefore, the enthalpy of formation of (Mn) and (Al) is equal to zero.
Now, put all the values in above formula, we get
![\Delta (H_{f})_{reaction}=[2moles\times (-1699.8 KJ/mole)}+3moles\times (0\text{ KJ/mole}})]-[(3moles\times(-520.0KJ/mole }+4moles\times(0\text{ KJ/mole})]](https://tex.z-dn.net/?f=%5CDelta%20%28H_%7Bf%7D%29_%7Breaction%7D%3D%5B2moles%5Ctimes%20%28-1699.8%20KJ%2Fmole%29%7D%2B3moles%5Ctimes%20%280%5Ctext%7B%20KJ%2Fmole%7D%7D%29%5D-%5B%283moles%5Ctimes%28-520.0KJ%2Fmole%20%7D%2B4moles%5Ctimes%280%5Ctext%7B%20KJ%2Fmole%7D%29%5D)
= (-3399.6) + (1560)
= -1839.6 KJ