Answer:
H₂S + Cl₂ —> S + 2HCl
Explanation:
? + Cl₂ —> S + 2HCl
To balance the equation above, we must recognise what atoms are present in the products.
The products contains S, H and Cl.
Thus, S, H and Cl must also be present in the reactants.
Considering the equation given above, we can see clearly that H and S is missing in the reactants.
H and S together as a compound is expressed as H₂S.
Now, we shall input H₂S into the equation to obtain the complete equation. This is illustrated below:
? + Cl₂ —> S + 2HCl
H₂S + Cl₂ —> S + 2HCl
Next, we shall verify to see if the equation is balanced.
There are 2 atoms of H on both sides of the equation.
There are 2 atoms of Cl on both sides of the equation.
1 atom of S exist on both sides of the equation.
Thus, the equation is balanced.
Hey there!
Molar mass Na2SO4 = <span>142.04 g/mol
Number of moles:
n = m / mm
n = 25 / 142.04
n = 0.176 moles of Na2SO4
Therefore, </span>use the Avogadro constant
1 mole Na2SO4 ------------------- 6.02x10²³ molecules
0.176 moles Na2SO4 ------------ molecules ??
0.176 x ( 6.02x10²³ ) / 1
=> 1.059x10²³ molecules of Na2SO4
hope this helps!
Can I see the article and what your researching
3.84 - 1.43 = 2.41
2.41g of table sugar
% mass = ( (mass of element) / (total mass) ) * 100
% mass = (2.41 / 3.84) * 100
% mass = (0.6276) * 100
% mass = 62.76
62.76%
Answer:
7.5 g of AlCl3
Explanation:
The given equation is;
NaOH + AlCl3 --> Al(OH)3 + NaCI.
By inspection, it is not balanced because OH and Clare not equal on both sides of the equation.
Thus, let's make them equal by balancing the equation.
Cl has 3 on the left, so we will make it to have 3 on the right. Same thing with OH on the right and we will make it to have 3 on the left. Thus:
3NaOH + AlCl3 --> Al(OH)3 + 3NaCI
We can see that;
NaOH has 3 moles
While AlCl3 has 1 mole
Thus, to find how many grams of AlCl3 will be required to completely react with 2.25g of NaOH ;
2.25g of NaOH × (3 moles NaOH/39.997 g/mol of NaOH) × (1 mole of AlCl3/3 moles of NaOH) × (133.34 g/mol of AlCl3/1 mol AlCl3) = 7.5 g of AlCl3