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riadik2000 [5.3K]
4 years ago
10

Can someone solve the equations for these? 45 points

Chemistry
1 answer:
Volgvan4 years ago
8 0
2)

formula equation: Pb(NO₃)₂(aq) + 2KI(aq) ----> PbI₂(s) + 2KNO₃(aq)
total ionic equation: Pb⁺² + 2NO₃⁻¹ + 2K⁺¹ + 2I⁻¹----> PbI₂(s) + 2K⁺¹ + 2NO₃⁻¹
net ionic equation: Pb⁺² + 2 I⁻¹----> PbI₂(s)

3) 

formula equation: Zn(NO₃)₂(aq) + K₂CO₃(aq) -----> ZnCO₃(s) + 2KNO₃(aq)
total ionic equation: Zn⁺² + 2NO₃⁻¹ + 2K⁺¹ + CO₃⁻² ---> ZnCO₃ (s) + 2K⁺¹ + 2NO₃⁻¹
net ionic equation: Zn⁺² + CO₃⁻² ----> ZnCO₃ (s)


note: if I did not specify the state of the molecule in the reaction, you can assume they are aqueous unless state otherwise. 
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Explanation:

The balanced equation of the reaction is given as;

Mg(OH)2 (s) + 2 HBr (aq) → MgBr2 (aq) + 2 H2O (l)

1. How many grams of MgBr2 will be produced from 18.3 grams of HBr?

From the reaction;

2 mol of HBr produces 1 mol of  MgBr2

Converting to masses using;

Mass = Number of moles * Molar mass

Molar mass of HBr = 80.91 g/mol

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(2 * 80.91 = 161.82g) of HBr produces (1 * 184.113 = 184.113g) MgBr2

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18.3 = x

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Converting the mass to mol;

Number of moles = Mass / Molar mass = 18.3 / 80.91 = 0.226 mol

From the reaction;

2 mol of HBr produces 2 mol of H2O

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3. How many grams of Mg(OH)2 are needed to completely react with 18.3 grams of HBr?

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2 mol of HBr reacts with 1 mol of Mg(OH)2

18.3g of HBr =  0.226 mol

2 = 1

0.226 = x

x = 0.226 * 1 /2

x = 0.113 mol

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