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irina1246 [14]
3 years ago
6

What is the correct net ionic equation, including all coefficients, charges, and phases, for the following set of reactants? Ass

ume that the contribution of protons from H2SO4 is near 100 %.Ba(OH)2(aq)+H2SO4(aq)→Express your answer as a chemical equation.
Chemistry
1 answer:
Nataly [62]3 years ago
4 0

Answer:

Ba^2+(aq) + 2OH-(aq) + 2H+(aq) + SO4^2-(aq) → BaSO4(s) + 2H2O(l)

Explanation:

Step 1: Data given

the contribution of protons from H2SO4 is near 100 %.

Step 2: The unbalanced equation

.Ba(OH)2(aq) + H2SO4(aq) → BaSO4(s) + H2O(l)

Step 3: Balancing the equation

On the left side we have 4x H (2x in Ba(OH)2 and 2x in H2SO4). On the right side, we have 2x H (in H2O).

To balance the amount of H on both sides, we have to muliply H2O (on the right side) by 2.

Ba(OH)2(aq) + H2SO4(aq) → BaSO4(s) + 2H2O(l)

Step 4: The net ionic equation

The net ionic equation, for which spectator ions are omitted - remember that spectator ions are those ions located on both sides of the equation - will.

Ba^2+(aq) + 2OH-(aq) + 2H+(aq) + SO4^2-(aq) → BaSO4(s) + 2H2O(l)

After canceling those spectator ions in both side, look like this:

Ba^2+(aq) + 2OH-(aq) + 2H+(aq) + SO4^2-(aq) → BaSO4(s) + 2H2O(l)

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How many moles of sodium carbonate are contained by 57.3g of sodium carbonate
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Answer:

\boxed {\boxed {\sf 0.541 \  mol \ Na_2CO_3}}

Explanation:

We are asked to find how many moles of sodium carbonate are in 57.3 grams of the substance.

Carbonate is CO₃ and has an oxidation number of -2. Sodium is Na and has an oxidation number of +1. There must be 2 moles of sodium so the charge of the sodium balances the charge of the carbonate. The formula is Na₂CO₃.

We will convert grams to moles using the molar mass or the mass of 1 mole of a substance. They are found on the Periodic Table as the atomic masses, but the units are grams per mole instead of atomic mass units. Look up the molar masses of the individual elements.

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Remember the formula contains subscripts. There are multiple moles of some elements in 1 mole of the compound. We multiply the element's molar mass by the subscript after it, then add everything together.

  • Na₂ = 22.9897693 * 2= 45.9795386 g/mol
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We will convert using dimensional analysis. Set up a ratio using the molar mass.

\frac {105.9875386  \ g \ Na_2CO_3}{1 \ mol \ Na_2CO_3}

We are converting 57.3 grams to moles, so we multiply by this value.

57.3 \ g \ Na_2CO_3} *\frac {105.9875386  \ g \ Na_2CO_3}{1 \ mol \ Na_2CO_3}

Flip the ratio so the units of grams of sodium carbonate cancel.

57.3 \ g \ Na_2CO_3} *\frac {1 \ mol \ Na_2CO_3}{105.9875386  \ g \ Na_2CO_3}

57.3 } *\frac {1 \ mol \ Na_2CO_3}{105.9875386 }

\frac {57.3 }{105.9875386 } \ mol \ Na_2CO_3

0.5406295944 \ mol \ Na_2CO_3

The original measurement of moles has 3 significant figures, so our answer must have the same. For the number we found that is the thousandth place. The 6 in the ten-thousandth place to the right tells us to round the 0 up to a 1.

0.541 \  mol \ Na_2CO_3

There are approximately <u>0.541 moles of sodium carbonate</u> in 57.3 grams.

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