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Lilit [14]
3 years ago
11

SiCl4 + H2O = H4SiO4 + HCl

Chemistry
1 answer:
vichka [17]3 years ago
8 0

Hey there!

Balance the equation:

SiCl₄ + H₂O → H₄SiO₄ + HCl

Balance H.

2 on the left, 5 on the right. Add a coefficient of 3 in front of H₂O and a coefficient of 2 in front of HCl.

SiCl₄ + 3H₂O → H₄SiO₄ + 2HCl  

Balance O.

3 on the left, 4 on the right. Change the coefficient of 3 in front of H₂O to a 4.

SiCl₄ + 4H₂O → H₄SiO₄ + 2HCl  

This unbalanced our H, so change the coefficient of 2 in front of HCl to a 4.

SiCl₄ + 4H₂O → H₄SiO₄ + 4HCl  

Balance Cl.

4 on the left, 4 on the right. Already balanced.

Balance Si.

1 on the left, 1 on the right. Already balanced.

Our final balanced equation:

SiCl₄ + 4H₂O → H₄SiO₄ + 4HCl

Hope this helps!

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In this reaction: Mg (s) + I₂ (s) → MgI₂ (s) If 2.68 moles of Mg react with 3.56 moles of I₂, and 1.76 moles of MgI₂ form, what
melomori [17]

Answer:

Y=65.7\%

Explanation:

Hello,

In this case, for the given chemical reaction, we first identify the limiting reactant by noticing that due to the 1:1 mole ratio for magnesium to iodine the reacting moles must the same, nevertheless, there are only 2.68 moles of magnesium versus 3.56 moles of iodine, for that reason, magnesium is the limiting reactant, so the theoretical turns out:

n_{MgI_2}^{theoretical}=2.68molMg*\frac{1molMgI_2}{1molMg} =2.68molMgI_2

Thus, we compute the percent yield as:

Y=\frac{n_{MgI_2}^{real}}{n_{MgI_2}^{theoretical}} *100\%=\frac{1.76mol}{2.68mol} *100\%\\\\Y=65.7\%

Best regards.

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hope this helps!

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