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!
Answer:
C
Explanation:
CH₄ is the formula for methane
When 0.34 of HNO₃ is titrated to equivalence using 0.14 l of 0.1 m NaOH then the concentration of HNO₃ is 0.041 M
The reaction of neutralization of HNO₃ with NaOH is
HNO₃ + NaOH → H₂O + NaNo₃
When 1 mole of HNO₃ react with 1 mole of NaOH, based on chemical rection the moles of NaOH at equivalence point are equal to moles of HNO₃ present in solution: -
With the mole and volumes, we can find molarity as follows:
Moles of NaOH = moles HNO₃
⁼ 0.14 L X (0.1 mol NaOH/L) = 0.014 mole NaOH
=0.014 mol HNO₃
Molarity: -
= 0.041 M
Thus, from above solution we concluded that the concentration of HNO₃ solution is 0.041 M.
Learn more about molarity: brainly.com/question/8732513
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No elements visible!
Ions form between metals and non-metals.
Hope this helps!
Answer:
[OH⁻] = 1.77 × 10⁻¹¹ M
Solution:
In order to calculate concentration of hydroxide ion i.e. [OH⁻] first we will find the pOH by using following relation,
pH + pOH = 14
Putting value of pH given and solving for pOH,
3.25 + pOH = 14
pOH = 14 - 3.25
pOH = 10.75
As ,
pOH = -log [OH⁻]
So,
[OH⁻] = 10^-pOH
Putting value of pOH,
[OH⁻] = 10⁻¹⁰·⁷⁵
[OH⁻] = 1.77 × 10⁻¹¹ M
