Explanation:
<h2>A. Calculating the molar mass of the ending compound.</h2>
In NH4OH, the compounds that make it up are NH4+ and OH-
Therefore N exists in the ammonium form.
In the ammonium ion 4H atoms are connected to N.
N is more electronegative than H, therefore when H bonds to N, H is the more positive atom therefore each H has a charge of +1, since there are 4 H atoms the charge contributed by the 4H atoms are +1 * 4 = +4
the overall charge of NH4 is +1
Charge of N (+) +4 = +1
Charge of N = +1 - 4
Therefore oxidation state of N in NH4 is = -3
Answer:
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Explanation:
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Answer:
92.4 grams.
Explanation:
- From the balanced reaction:
<em>CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O,</em>
1.0 mole of CaCO₃ reacts with 2.0 moles of HCl to produce 1.0 mole of CaCl₂, 1.0 mole of CO₂, and 1.0 mole of H₂O.
- We need to calculate the no. of moles of (104 g) of CaCO₃:
<em>no. of moles of CaCO₃ = mass/molar mass</em> = (104 g)/(100.08 g/mol) = <em>1.039 mol.</em>
<u><em>Using cross multiplication:</em></u>
1.0 mole of CaCO₃ produce → 1.0 mole of CaCl₂.
∴ 1.039 mole of CaCO₃ produce → 1.039 mole of CaCl₂.
∴ The amount of CaCl₂ produced = no. of moles x molar mass = (1.039 mol)(110.98 g/mol) = 114.3 g.
∵ percent yield of the reaction = [(actual yield)/(theoretical yield)] x 100.
Percent yield of the reaction = 80.15%, theoretical yield = 115.3 g.
<em>∴ actual yield = [(percent yield of the reaction)(theoretical yield)]/100 </em>= [(80.15%)/(115.3 g)] / 100 = <em>92.42 g ≅ 92.4 g.</em>
