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Alexxandr [17]
3 years ago
5

Write net ionic equations for gas-forming reactions. Write a net ionic equation for the reaction that occurs when solid barium c

arbonate is combined with excess aqueous hydroiodic acid.
Chemistry
1 answer:
AysviL [449]3 years ago
7 0

Answer: BaCO3(s) + 2H+(aq) --> Ba+2 + H2O(l) + CO2(g)

Explanation: Net ionic equations show only the species or atoms that participate in the reaction, in this case gas-forming

When solid barium carbonate is combined with excess aqueous hydroiodic acid the reaction that take place is.

BaCO3(s) + 2HI(aq) --> BaI2 (aq) + H2O(l) + CO2(g)

To write a ionic equation is necesary to separte the species that are dissociated in water (ag) and expresed it as ions showing the charge

BaCO3(s) + 2I-(aq) + 2H+(aq) --> Ba+2 + 2I- (aq) + H2O(l) + CO2(g)

Since I- is present in both sides of the reaction, it's eliminated in the net ionic equation, and the final expression is:

BaCO3(s) + 2H+(aq) --> Ba+2 + H2O(l) + CO2(g)

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1) When 2.38g of magnesium is added to 25.0cm of 2.27 M hydrochloric acid, hydrogen gas is released.
Andrej [43]

Answer:

a. HCl.

b. 0.057 g.

c. 1.69 g.

d. 77 %.

Explanation:

Hello!

In this case, since the reaction between magnesium and hydrochloric acid is:

Mg+2HCl\rightarrow MgCl_2+H_2

Whereas there is 1:2 mole ratio between them.

a) Here, we can identify the limiting reactant as that yielded the fewest moles of hydrogen gas product via the 1:1 and 2:1 mole ratios:

n_{H_2}^{by\  HCl}=0.025L*2.27\frac{molHCl}{1L}*\frac{1molH_2}{2molHCl}  =0.0284molH_2\\\\n_{H_2}^{by\  Mg}=2.38gMg*\frac{1molMg}{24.3gMg}*\frac{1molH_2}{1molMg}=0.0979molH_2

Thus, since hydrochloric yields fewer moles of hydrogen than magnesium, we realize it is the limiting reactant.

b) Here, we use the molar mass of gaseous hydrogen (2.02 g/mol) to compute the mass:

m_{H_2}=0.0284molH_2*\frac{2.02gH_2}{1molH_2}=0.057gH_2

c) Here, we compute the mass of magnesium associated with the yielded 0.0248 moles of hydrogen:

m_{Mg}^{reacted}=0.0284molH_2*\frac{1molMg}{1molH_2}*\frac{24.3gMg}{1molMg}  =0.690gMg

Thus, the mass of excess magnesium turns out:

m_{Mg}^{excess}=2.38g-0.690g=1.69gMg

d) Finally, we compute the percent yield, considering 0.044 g is the actual yield and 0.057 g the theoretical yield:

Y=\frac{0.044g}{0.057g} *100\%\\\\Y=77\%

Best regards!

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