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3 years ago

Balance the equation Mg+Fe2O3->MgO+Fr

Chemistry

2 answers:

Readme [11.4K]3 years ago

8 0

Answer:

3Mg +Fe2O3 ---> 3 MgO + 2Fe

Explanation:

3Mg +Fe2O3 ---> 3 MgO + 2Fe

Mg 3 Mg 3

Fe 2 Fe 2

O 3 O 3

before after

sergij07 [2.7K]3 years ago

4 0

Answer:

3Mg + Fe2O3 = 2Fe + 3MgO

Explanation:

Make sure the number of each element equal on both right and left hand side

Eg. There are 3 molecules of oxygen on left hand side so to balance right hand side add a 3 in front of MgO, because you've added a 3 in front of MgO you now have to add a 3 in front of Mg on the left hand side to balance it out and continue to do so for the rest of the chemical equation.

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If 20.0 g of NaOH is added to 0.750 L of 1.00 M Cd(NO₃)₂, how many grams of Cd(OH)₂ will be formed in the following precipitatio

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Answer:

$m_{Cd(OH)_2}=36.6 gCd(OH)_2$

Explanation:

Hello.

In this case, for the given chemical reaction, in order to compute the grams of cadmium hydroxide that would be yielded, we must first identify the limiting reactant by computing the yielded moles of that same product, by 20.0 grams of NaOH (molar mass = 40 g/mol) and by 0.750 L of the 1.00-M solution of cadmium nitrate as shown below considering the 1:2:1 mole ratios respectively:

$n_{Cd(OH)_2}^{by\ NaOH}=20.0gNaOH*\frac{1molNaOH}{40gNaOH} *\frac{1molCd(OH)_2}{2molNaOH} =0.25molCd(OH)_2\\\\n_{Cd(OH)_2}^{by\ Cd(NO_3)_2}=0.750L*1.00\frac{molCd(NO_3)_2}{L}*\frac{1molCd(OH)_2}{1molCd(NO_3)_2} =0.75molCd(OH)_2$

Thus, since 20.0 grams of NaOH yielded less of moles of cadmium hydroxide, NaOH is the limiting reactant, therefore the mass of cadmium hydroxide (molar mass = 146.4 g/mol) is:

$m_{Cd(OH)_2}=0.25molCd(OH)_2*\frac{146.4gCd(OH)_2}{1molCd(OH)_2} \\\\m_{Cd(OH)_2}=36.6 gCd(OH)_2$

Best regards.

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3 years ago