Question #5715d

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Answer 1 · 2015-07-27T15:59:09

(a). $2 \times 10^{22}$ $2xx10^22$

(b). $4 \times 10^{22}$ $4xx10^22$

(c). $2 \times 10^{22}$ $2xx10^22$

(d). $66 mg$ $66"mg"$

(a).

$C u^{2 +} + 2 e \to C u$ $Cu^(2+)+2erarr Cu$

Mass Cu = 2.12g

So moles Cu = $\frac{2.12}{63.6} = 0.033$ $2.12/63.6=0.033$

So no. atoms Cu = $0.033 \times 6.02 \times 10^{23} = 2.0 \times 10^{22}$ $0.033xx6.02xx10^(23)=2.0xx10^22$

(b).

From the above equation 1 mole $C u^{2 +}$ $Cu^(2+)$ ions require 2 moles of electrons to be discharged.

So no. moles electrons $= 0.033 \times 2 = 0.066$ $=0.033xx2=0.066$

So no. electrons $= 0.066 \times 6.02 \times 10^{23} = 4 \times 10^{22}$ $= 0.066xx6.02xx10^(23)=4xx10^22$

(c).

$2 H^{+} + 2 e \to H_{2}$ $2H^(+)+2erarr H_2$

So you can see a mole of hydrogen gas also requires 2 moles of electrons to be discharged.

so no. molecules $H_{2} = 2 \times 10^{22}$ $H_2=2xx10^22$ .

(d).

No. moles $H_{2}$ $H_2$ = no. moles $C u = 0.033$ $Cu =0.033$

$M_{r} H_{2} = 2.0$ $M_rH_2=2.0$

So mass $H_{2} = 0.033 \times 2 = 0.066 g = 66 mg$ $H_2=0.033xx2=0.066"g"=66"mg"$