Answer:
1. The products of this reaction are ZnCl₂ and H₃PO₄.
2. 14.57 g.
Explanation:
<em>1. What would the products of this reaction be?</em>
- The balanced reaction between Zn₃(PO₄)₂ and HCl is represented as:
<em>Zn₃(PO₄)₂ + 6HCl → 3ZnCl₂ + 2H₃PO₄,</em>
It is clear that 1.0 mol of Zn₃(PO₄)₂ reacts with 6.0 mol of HCl to produce 3.0 mol of ZnCl₂ and 2.0 mol of H₃PO₄.
So, the products of this reaction are ZnCl₂ and H₃PO₄.
<em>2. If we produced 13.05 g of H₃PO₄, how many grams of hydrochloric acid would be need to start with?</em>
- Firstly, we should get the no. of moles (n) of 13.05 grams of H₃PO₄:
n = mass/molar mass = (13.05 g)/(97.994 g/mol) = 0.1332 mol.
<u><em>Using cross-multiplication:</em></u>
6.0 mol of HCl needed to produce → 2.0 mol of H₃PO₄, from stichiometry.
??? mol of HCl needed to produce → 0.1332 mol of H₃PO₄.
∴ The no. of moles of HCl needed = (6.0 mol)(0.1332 mol)/(2.0 mol) = 0.3995 mol.
∴ The mass of HCl needed = n*molar mass = (0.3995 mol)(36.46 g/mol) = 14.57 g.
<em>So, the grams of hydrochloric acid would be need to start with = 14.57 g.</em>
Hello there!
I believe "<span>b. Nutrients move from an area of lower concentration to an area of higher concentration." is correct!
Have a cool day! -Wajiha</span>
Answer:
<em><u>Copper </u></em><em><u>(</u></em><em><u>Cu2)</u></em><em><u> </u></em><em><u>,</u></em><em><u> </u></em><em><u>Iron </u></em><em><u>(</u></em><em><u>Fe2+</u></em><em><u> </u></em><em><u>Fe3 </u></em><em><u>+</u></em><em><u>)</u></em><em><u> </u></em><em><u>,</u></em><em><u> </u></em><em><u>and </u></em><em><u>Hydrogen </u></em><em><u>ion </u></em><em><u>(</u></em><em><u>H+</u></em><em><u>)</u></em>
Explanation:
I hope it helps u dear! ^_^
You're looking for the number of moles of H2, and you have 6.0 mol Al and 13 mol HCL.
For the first part, you have to make your way from 6.0 mol of Al to mol of H2, right? For that to happen, you need to make a conversion factor that will cancel the mol Al, in such case use the 2 moles of Al from your equation to cancel them out. At the top of the equation, you can use the number of moles of H2 from the equation and find the moles that will be produced for the H2.
6.0mol Al x 3 mol H2/2 mol Al = 9 mol H2
For the second part, you have to make the same procedure, make a conversion factor that will cancel the mol of HCL and for that you need to use the 6 mol HCL from your equation, and at the numerator you can put the 3 mol of H2 from the equation so that you can find the number of moles of H2 that will be produced.
13 mol HCL x 3 mol H2/6 mol HCL = 6.5 mol H2
As it can be seen, HCL produces the less amount of H2 moles. Therefore, the reaction CANNOT produce more than 6.5 mol H2, in that case 6.5 mol will be the maximum number of moles that will be produced at the end because HCL does not have enough to produce more than 6.5 mol.
In that case HCL is the limiting reactant because it limits that will be produced, and so the answer is B!