Answer:
191.6 g of CaCl₂.
Explanation:
What is given?
Mass of HCl = 125.9 g.
Molar mass of CaCl₂ = 110.8 g/mol.
Molar mass of HCl = 36.4 g/mol.
Step-by-step solution:
First, we have to state the chemical equation. Ca(OH)₂ react with HCl to produce CaCl₂:
Now, let's convert 125.9 g of HCl to moles using the given molar mass (remember that the molar mass of a compound can be found using the periodic table). The conversion will look like this:
Let's find how many moles of CaCl₂ are being produced by 3.459 moles of HCl. You can see in the chemical equation that 2 moles of HCl reacted with excess Ca(OH)₂ produces 1 mol of CaCl₂, so we state a rule of three and the calculation is:
The final step is to find the mass of CaCl₂ using the molar mass of CaCl₂. This conversion will look like this:
The answer would be that we're producing a mass of 191.6 g of CaCl₂.
Answer:
FeCl₃
Explanation:
4FeCl₃ + 3O₂ => 2Fe₂O₃+ 6Cl₂
Given => 7moles 9moles
A simple way to determine which reagent is the limiting reactant is to convert all given data to moles then divide by the respective coefficients of the balanced equation. The smaller value will be the limiting reactant.
4FeCl₃ + 3O₂ => 2Fe₂O₃+ 6Cl₂
Given => 7/4 = 1.75* 9/3 = 3
*Smaller value => FeCl₃ is limiting reactant.
NOTE: However, when working problems, one must use original mole values given.
Answer:
1.403x10²⁴ molecules
Explanation:
In order to calculate how many molecules of CO₂ are there in 102.5 g of the compound, we first<u> convert grams to moles</u> using its <em>molar mass</em>:
- 102.5 g ÷ 44 g/mol = 2.330 mol CO₂
Now we <u>convert moles into molecules </u>using <em>Avogadro's number</em>:
- 2.330 mol * 6.023x10²³ molecules/mol = 1.403x10²⁴ molecules
E.g. in H3PO4 (O, -2).
8. The sum of the oxidation states of all the atoms in a species must be equal to net charge on the species. e.g. Net Charge of HClO4 = 0, i.e. [+1(H)+7(Cl)-2<span>*4(O)] = 0.</span>
