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Chemistry

1 answer:

lara [203]3 years ago

3 0

Pb + Mg(NO₃)₂ → Pb(NO₃)₂ + Mg
This reaction would NOT occur because Pb is less reactive than Mg and as such Pb cannot displace the Mg in order for the reaction to occur under normal conditions.

Mg + Fe(NO₃)₂ → Fe + Mg(NO₃)₂
This reaction would occur. This is because Mg is more reactive than Fe and as such can displace it in the reaction, thus allowing the reaction to occur under normal conditions.

Cu + Mg(NO₃)₂ → Cu(NO₃)₂ + Mg
This reaction would NOT occur. Mg is more reactive than Cu, and as such copper cannot displace magnesium in order for the reaction to occur under normal conditions.

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2 years ago

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

bulgar [2K]

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:

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

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Answer:

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