The balanced chemical equation for the above reaction is as follows ;
Mg + 2HCl —> MgCl2 + H2
The stoichiometry of Mg to HCl is 1:2
This means that 1 mol of Mg reacts with 2 mol of HCl
Equal amounts of both Mg and HCl have been added. One reagent is the limiting reactant and other reactant is in excess.
Limiting reactant is the reagent that is fully used up in the reaction and the amount of Product formed depends on the amount of limiting reactant present.
In this reaction if Mg is the limiting reactant, 4.40 moles of Mg should react with 4.40x2 -8.80 moles of HCl.
But only 4.40 moles of HCl present therefore HCl is the limiting reactant that reacts with 4.40/2 = 2.20 moles of Mg
Stoichiometry of HCl to MgCl2 is 2:1
Since HCl moles reacted -4.40 mol
Then MgCl2 moles formed are 4.40/2 = 2.20 mol of MgCl2
<span><span>Standard Reduction Potentials in Aqueous Solution at 25oC</span><span>Acidic Solution<span>Eo (V)</span></span><span>F2(g) + 2 e– 2 F–(aq)+2.87</span><span>Co3+(aq) + e– Co2+(aq)+1.82</span><span>Pb4+(aq) + 2 e– Pb2+(aq)+1.8</span><span>H2O2(aq) + 2 H+(aq) + 2 e– 2 H2O+1.77</span><span>NiO2(s) + 4 H+(aq) + 2 e– Ni2+ (aq) + 2 H2O+1.7</span><span>PbO2(s) + SO42–(aq) + 4 H+(aq) + 2 e– PbSO4(s) + 2 H2O+1.685</span><span>Au+(aq) + e– Au(s)+1.68</span><span>2 HClO(aq) + 2 H+(aq) + 2 e– Cl2(g) + 2 H2O+1.63</span><span>Ce4+(aq) + e– Ce3+(aq)+1.61</span><span>NaBiO3(s) + 6 H+(aq) + 2 e– Bi3+(aq) + Na+(aq) + 3 H2O+~1.6</span><span>MnO4–(aq) + 8 H+(aq) + 5 e– Mn2+(aq) + 4 H2O+1.51</span><span>Au3+(aq) + 3 e– Au(s)+1.5</span><span>ClO3–(aq) + 6 H+(aq) + 5 e– 1/2 Cl2(g) + 3 H2O+1.47</span><span>BrO3– + 6 H+(aq) + 6 e– Br–(aq) + 3 H2O+1.44</span><span>Cl2(g) + 2 e– 2 Cl–(aq)+1.358</span><span>Cr2O72– + 14 H+(aq) + 6 e– 2 Cr3+(aq) + 7 H2O+1.33</span><span>N2H5+(aq) + 3 H+(aq) + 2 e– 2 NH4+(aq)+1.24</span><span>MnO2(s) + 4 H+(aq) + 2 e– Mn2+(aq) + 2 H2O+1.23</span><span>O2(g) + 4 H+(aq) + 4 e– 2 H2O+1.229</span><span>Pt2+(aq) + 2 e– Pt(s)+1.2</span><span>IO3–(aq) + 6 H+(aq) + 5 e– 1/2 I2(aq) + 3 H2O+1.195</span><span>ClO4–(aq) + 2 H+(aq) + 2 e– ClO3–(aq) + H2O+1.19</span><span>Br2(l) + 2 e– 2 Br–(aq)+1.066</span><span>AuCl4– + 3 e– Au(s) + 4 Cl–(aq)+1</span><span>Pd2+(aq) + 2 e– Pd(s)+0.987</span><span>NO3–(aq) + 4 H+(aq) + 3 e<span>– </span> NO(g) + 2 H2O+0.96</span><span>NO3–(aq) + 3 H+(aq) + 2 e– HNO2(aq) + H2O+0.94</span><span>2 Hg2+(aq) + 2 e– Hg22+(aq)+0.92</span><span>Hg2+(aq) + 2 e– Hg(l)+0.855</span><span>Ag+(aq) + e– Ag(s)+0.7994</span><span>Hg22+(aq) + 2 e– 2 Hg(l)+0.789</span><span>Fe3+(aq) + e– Fe2+(aq)+0.771</span><span>SbCl6–(aq) + 2 e– SbCl4–(aq) + 2 Cl–(aq)+0.75</span><span>[PtCl4]2–(aq) + 2 e– Pt(s) + 4 Cl–(aq)+0.73</span><span>O2(g) + 2 H+(aq) + 2 e– H2O2(aq)+0.682</span><span>[PtCl6]2–(aq) + 2 e– [PtCl4]2–(aq) + 2 Cl–(aq)+0.68</span><span>H3AsO4(aq) + 2 H+(aq) + 2 e– H3AsO3(aq) + H2O+0.58</span><span>I2(s) + 2 e– 2 I–(aq)+0.535</span><span>TeO2(s) + 4 H+(aq) + 4 e– Te(s) + 2 H2O+0.529</span><span>Cu+(aq) + e– Cu(s)+0.521</span><span>[RhCl6]3–(aq) + 3 e– Rh(s) + 6 Cl–(aq)+0.44</span><span>Cu2+(aq) + 2 e– Cu(s)+0.337</span><span>HgCl2(s) + 2 e– 2 Hg(l) + 2 Cl–(aq)+0.27</span><span>AgCl(s) + e– Ag(s) + Cl–(aq)+0.222</span><span>SO42–(aq) + 4 H+(aq) + 2 e– SO2(g) + 2 H2O+0.2</span><span>SO42–(aq) + 4 H+(aq) + 2 e– H2SO3(g) + H2O+0.17</span><span>Cu2+(aq) + e– Cu+(aq)+0.153</span><span>Sn4+(aq) + 2 e– Sn2+(aq)+0.15</span><span>S(s) + 2 H+(aq) + 2 e– H2S(aq)+0.14</span><span>AgBr(s) + e– Ag(s) + Br–(aq)+0.0713</span><span>2 H+(aq) + 2 e– H2(g) (reference electrode)0</span><span>N2O(g) + 6 H+(aq) + H2O + 4 e– 2 NH3OH+(aq)-0.05</span><span>Pb2+(aq) + 2 e– Pb(s)-0.126</span><span>Sn2+(aq) + 2 e– Sn(s)-0.14</span><span>AgI(s) + e– Ag(s) + I–(aq)-0.15</span><span>[SnF6]2–(aq) + 4 e– Sn(s) + 6 F–(aq)-0.25</span><span>Ni2+(aq) + 2 e– Ni(s)-0.25</span><span>Co2+(aq) + 2 e– Co(s)-0.28</span><span>Tl+(aq) + e– Tl(s)-0.34</span><span>PbSO4(s) + 2 e– Pb(s) + SO42–(aq)-0.356</span><span>Se(s) + 2 H+(aq) + 2 e<span>– </span> H2Se(aq)-0.4</span><span>Cd2+(aq) + 2 e– Cd(s)-0.403</span><span>Cr3+(aq) + e– Cr2+(aq)-0.41</span><span>Fe2+(aq) + 2 e– Fe(s)-0.44</span><span>2 CO2(g) + 2 H+(aq) + 2 e– (COOH)2(aq)-0.49</span><span>Ga3+(aq) + 3 e– Ga(s)-0.53</span><span>HgS(s) + 2 H+(aq) + 2 e– Hg(l) + H2S(g)-0.72</span><span>Cr3+(aq) + 3 e– Cr(s)-0.74</span><span>Zn2+(aq) + 2 e– Zn(s)-0.763</span><span>2H2O(l) + 2 e– H2(g) + 2OH-(aq)-0.8277</span><span>Cr2+(aq) + 2 e<span>– </span> Cr(s)-0.91</span><span>Mn2+(aq) + 2 e– Mn(s)-1.18</span><span>V2+(aq) + 2 e– V(s)-1.18</span><span>Zr4+(aq) + 4 e– Zr(s)-1.53</span><span>Al3+(aq) + 3 e– Al(s)-1.66</span><span>H2(g) + 2 e– 2 H–(aq)-2.25</span><span>Mg2+(aq) + 2 e– Mg(s)-2.37</span><span>Na+(aq) + e– Na(s)-2.714</span><span>Ca2+(aq) + 2 e– Ca(s)-2.87</span><span>Sr2+(aq) + 2 e– Sr(s)-2.89</span><span>Ba2+(aq) + 2 e<span>– </span> Ba(s)-2.9</span><span>Rb+(aq) + e– Rb(s)-2.925</span><span>K+(aq) + e– K(s)-2.925</span><span>Li+(aq) + e– Li(s)-3.045</span><span>Basic Solution </span><span>ClO–(aq) + H2O + 2 e– Cl–(aq) + 2 OH–(aq)0.89</span><span>OOH–(aq) + H2O + 2 e– 3 OH–(aq)0.88</span><span>2 NH2OH(aq) + 2 e– N2H4(aq) + 2 OH–(aq)0.74</span><span>ClO3– (aq) + 3 H2O + 6 e– Cl–(aq) + 6 OH–(aq)0.62</span><span>MnO4–(aq) + 2 H2O + 3 e– MnO2(s) + 4 OH–(aq)0.588</span><span>MnO4–(aq) + e– MnO42–(aq)0.564</span><span>NiO2(s) + 2 H2O + 2 e- Ni(OH)2(s) + 2 OH–(aq)0.49</span><span>Ag2CrO4(s) + 2 e– 2 Ag(s) + CrO42–(aq)0.446</span><span>O2(g) + 2 H2O + 4 e– 4 OH–(aq)0.4</span><span>ClO4–(aq) + H2O + 2e– ClO3–(aq) + 2 OH–(aq)0.36</span><span>Ag2O(s) + H2O + 2e– 2 Ag(s) + 2 OH–(aq)0.34</span><span>2 NO2–(aq) + 3 H2O + 4 e– N2O(g) + 6 OH–(aq)0.15</span><span>N2H4(aq) + 2 H2O + 2 e– 2 NH3(aq) + 2 OH–(aq)0.1</span><span>[Co(NH3)6]3+(aq) + e– [Co(NH3)6]2+(aq)0.1</span><span>HgO(s) + H2O + 2e– Hg(l) + 2 OH–(aq)0.0984</span><span>O2(g) + H2O + 2 e– OOH–(aq) + OH–(aq)0.076</span><span>NO3–(aq) + H2O + 2 e– NO2–(aq) + 2 OH–(aq)0.01</span><span>MnO2(s) + 2 H2O + 2 e– Mn(OH)2(s) + 2 OH–(aq)-0.05</span><span>CrO42–(aq) + 4 H2O + 3 e– Cr(OH)3(s) + 5 OH–(aq)-0.12</span><span>Cu(OH)2(s) + 2 e– Cu(s) + 2 OH–(aq)-0.36</span><span>Fe(OH)3(s) + e– Fe(OH)2(s) + OH–(aq)-0.56</span><span>2 H2O + 2 e– H2(g) + 2 OH–(aq)-0.8277</span><span>2 NO3–(aq) + 2 H2O + 2 e– N2O4(g) + 4 OH–(aq)-0.85</span><span>Fe(OH)2(s) + 2 e– Fe(s) + 2 OH–(aq)-0.877</span><span>SO42–(aq) + H2O + 2 e– SO32–(aq) + 2 OH–(aq)-0.93</span><span>N2(g) + 4 H2O + 4 e– N2H4(aq) + 4 OH–(aq)-1.15</span><span>[Zn(OH)4]2–(aq) + 2 e– Zn(s) + 4 OH–(aq)-1.22</span><span>Zn(OH)2(s) + 2 e– Zn(s) + 2 OH–(aq)-1.245</span><span>[Zn(CN)4]2–(aq) + 2 e– Zn(s) + 4 CN–(aq)-1.26</span><span>Cr(OH)3(s) + 3 e– Cr(s) + 3 OH–(aq)-1.3</span><span>SiO32–(aq) + 3 H2O + 4 e– Si(s) + 6 OH–(aq)<span>-1.7
try to understand</span></span></span>
Answer:
a) gold - Solid
b) gasoline - liquid
c) oxygen - gaseous
d) olive oil - liquid
e) mercury (found in thermometers) - liquid
f) aluminum - solid
Explanation:
Matter exists in three states viz: solid, gas and liquid. Note that, room temperature is a neither hot or cold temperature, which is about 20-23°C.
A) Gold is a transition metal with the symbol Au. At a room temperature, Gold, like most metals are found as SOLIDS.
B) Gasoline is a liquid mixture of refined hydrocarbons majorly used as fuels.
C) Oxygen is a chemical element with symbol O. It is a constituent of natural air, which is a mixture of different gases including oxygen. Hence, at room temperature, Oxygen is a gas.
D) Olive oil is a vegetable oil got from olives. It is composed of fats and it's a liquid at room temperature.
E) Mercury is an element with symbol Hg. It is the only metallic element found naturally (at room temperature) as a liquid. It is used in thermometers.
F) Aluminum is another chemical element with symbol Al. It is a lustrous metal that occurs as a Solid in room temperature.
Answer: Option C is not a property of a acid, acids turn blue litmus paper red.
Explanation:
In the malate–aspartate shuttle, electrons are transferred from <u>oxaloacetate</u> to form malate.
A crucial mechanism employed by mitochondria, the malate-aspartate shuttle system, also known as the malate shuttle, enables electrons to pass through the impermeable membrane separating the cytosol as well as the mitochondrial matrix. Glycolysis produces the electrons which are required for oxidative phosphorylation.
The liver, heart, as well as kidney, include the malate-aspartate shuttle, which produces about three molecules of ATP for every cytosolic NADH molecule. Under aerobic conditions, it constitutes quantitatively the most significant shuttle for such reoxidation of cytosolic NADH in vertebrate tissues.
Therefore, In the malate–aspartate shuttle, electrons are transferred from <u>oxaloacetate</u> to form malate.
To know more about electrons
brainly.com/question/1255220
#SPJ4