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ryzh [129]
3 years ago
15

If the reaction yield is 94.4%, what mass in grams of hydrogen is produced by the reaction of 4.73 g of magnesium with 1.83 g of

water? mg(s)+2h2o(l)→mg(oh)2(s)+h2(g)
Chemistry
1 answer:
CaHeK987 [17]3 years ago
6 0
First, it is important to list the molar mass of the relevant substances.

Molar mass of magnesium = 24.305 g/mol
Molar mass of water = 18.0153 g/mol
Molar mass of H2 = 2.0159 g/mol

Second, we need to determine the limiting reactant for the chemical reaction. We take 4.73 g of Mg and determine the stoichiometric amount of water needed for it to be completely consumed. This is shown in the following equation:

4.73 g Mg x mol Mg/24.305 g x 2 mol H2O/1 mol Mg x 18.0153 g/mol H2O = 7.0119 g H2O

Thus, we have determined that 7.0119 g H2O is needed to completely react 4.73 g Mg. The given amount of 1.83 g H2O is insufficient which then indicates that water is the limiting reactant and should be the basis of our calculations. 

Next, given 1.83 g H20, we calculate the theoretical yield of hydrogen gas using stoichiometry. The equation is then:

1.83 g H2O x mol H20/18.0153 g x 1 mol H2/2 mol H2O x 2.0159 g/mol H2 = 0.1024 g H2

However, the reaction yield was given to be 94.4%. The reaction yield is given by the formula percent yield = actual yield/ theoretical yield x 100%. Thus, the actual yield of hydrogen gas can be determined using the formula. 

Actual yield of H2 = 0.94*0.1024 g H2

Thus, the amount of hydrogen gas produced is 0.0963 g.   
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Ierofanga [76]

Answer:

K= 25.16%

Mn= 34.59%

O= 40.25%

Explanation:

KMnO_4

Find the total mass of the compound. You can do this by multiplying by the molar mass.

K=40*1=40\\Mn=55*1=55\\O=16*4=64

add them:

40+55+64=159g

Now convert each element individually to grams using molar mass

1molK(\frac{40gK}{1molK} )=40gK

1molMn(\frac{55gMn}{1molMn} )=55gMn

4molO(\frac{16gO}{1molO})=64gO

Divide by the mass of the compound and multiply by 100

K=\frac{40g}{159g} *100=25.16

Mn=\frac{55g}{159g}*100=34.59

O=\frac{64g}{159g}*100=40.25

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The given compound is copper (II) hydroxide, Cu(OH)_{2}. According to the solubility rules the hydroxides of all metal ions are insoluble except that of ammonium and the alkali metal ions. So copper(II) hydroxide will be insoluble in water. In aqueous solution, copper (II) hydroxide would exist as a sparingly soluble compound in which the undissociated compound remains in equilibrium with the ions Cu^{2+}and OH^{-}. The equilibrium would remain mostly towards undissociated form.

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