II. Pure magnesium metal is often found as ribbons and can easily burn in the presence of oxygen. When 3.86 g of magnesium ribbon burns with 155ml of oxygen at 1.0 atm and 275 K , a bright, white light and a white, powdery product is formed, magnesium oxide. How many grams of excess reactant remain?

Answers

Answer 1

Answer:

[tex]Excess=3.53g[/tex]

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

[tex]2Mg(s)+O_2(g) \rightarrow 2MgO(s)[/tex]

Next, we identify the limiting reactant by computing the moles of magnesium oxide yielded by 3.86 g of magnesium and 155 mL of oxygen at the given conditions via their 2:1:2 mole ratios and the ideal gas equation:

[tex]n_{MnO}^{by \ Mg}=3.86gMg*\frac{1molMg}{24.3gMg}*\frac{2molMgO}{2molMg} =0.159molMgO\\\\n_{MnO}^{by \ O_2}=\frac{1atm*0.155L}{0.082\frac{atm*L}{molO_2*K}*275K} *\frac{2mol MgO}{1molO_2} =0.0137molMgO[/tex]

It means that the limiting reactant is the oxygen as it yields the smallest amount of magnesium oxide. Next, we compute the mass of magnesium consumed the oxygen only:

[tex]m_{Mg}^{consumed}=0.0137molMgO*\frac{2molMg}{2molMgO} *\frac{24.3gMg}{1molMg} =0.334gMg[/tex]

Thus, the mass in excess is:

[tex]Excess=3.86g-0.334g\\\\Excess=3.53g[/tex]

Regards!


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Answers

Answer:

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Explanation:

Given data:

Mass of magnesium = 7.73 g

Mass of water = 1.31 g

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Mass of hydrogen produced = ?

Solution:

Chemical equation:

Mg + 2H₂O     →   Mg(OH)₂ + H₂

Number of moles of water:

Number of moles = mass/ molar mass

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Number of moles of magnesium:

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Now we will compare the moles of water and magnesium with hydrogen.

                    Mg            :           H₂

                     1               :            1

                    0.43          :          0.43

                    H₂O          :             H₂

                      2             :              1

                   0.07           :            1/2×0.07 = 0.035

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Answers

Answer:

6 grains

Explanation:

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Answers

Answer:

Option C: Use enough cations and anions so that the total charge is zero

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