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vfiekz [6]
3 years ago
5

How many moles of MgCl2 are there in 319 g of the compound?

Chemistry
1 answer:
Leokris [45]3 years ago
3 0

Hey there!:

Molar mass MgCl2 = 95.2110 g/mol

So:

1 mole MgCl2 -------------- 95.2110 g

moles MgCl2 ---------------- 319 g

moles MgCl2 = 319 * 1 / 95.2110

moles MgCl2 = 319 / 95.2110

=> 3.350 moles of MgCl2


Hope that helps!

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Answer : 0.0392 grams of Zn metal would be required to completely reduced the vanadium.

Explanation :

Let us rewrite the given equations again.

2VO_{2}^{+} (aq)+ 4H^{+}(aq) + Zn (s)\rightarrow 2VO^{2+}(aq)+Zn^{2+}+2H2O(l)2VO^{2+}(aq)+ 4H^{+}(aq) + Zn (s)\rightarrow 2V^{3+} (aq)+Zn^{2+}+2H2O(l)

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On adding above equations, we get the following combined equation.

2VO_{2}^{+} (aq)+ 8H^{+} (aq) + 3Zn (s)\rightarrow 2V^{2+}(aq)+3Zn^{2+}(aq)+4H_{2}O(l)

We have 12.1 mL of 0.033 M solution of VO₂⁺.

Let us find the moles of VO₂⁺ from this information.

12.1 mL \times \frac{1L}{1000mL}\times \frac{0.033mol}{L}=0.0003993mol NO_{2}^{+}

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Let us use this as a conversion factor to find the moles of Zn.

0.0003993mol NO_{2}^{+}\times \frac{3mol Zn}{2molNO_{2}^{+}}=0.00059895mol Zn

Let us convert the moles of Zn to grams of Zn using molar mass of Zn.

Molar mass of Zn is 65.38 g/mol.

0.00059895mol Zn\times \frac{65.38gZn}{1molZn}=0.0392gZn

We need 0.0392 grams of Zn metal to completely reduce vanadium.

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