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3 years ago

14

consider the balanced chemical equation below. when the chemical reaction was carried out calculated theoretical was yield for s

odium bromide 162 grams but the measured yield was 150 grams what is the percent yield?

1 answer:

8_murik_8 [283]3 years ago

7 0

Answer:

Explanation:

% yield = (actual yield / theoretical yield) X 100

For this question,

% yield = (150g/ 162 g) X 100 = 92.6%

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Molar mass of oxygen gas:

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Provide a balanced molecular equation, total ionic, and net ionic equation for sodium phosphate and zinc acetate.

vivado [14]

Answer: Balanced molecular equation :

$2Na_3PO_4(aq)+3(CH_3COO)_2Zn(aq)\rightarrow 6CH_3COONa(aq)+Zn_3(PO_4)_2(s)$

Total ionic equation:

$6Na^+(aq)+3PO_4^{2-}(aq)+6CH_3COO^-(aq)+3Zn^{2+}(aq)\rightarrow 6CH_3COO^-(aq)+6Na^+(aq)+Zn_3(PO_4)_2(s)$

The net ionic equation:

$2PO_4^{3-}(aq)+3Zn^{2+}(aq)\rightarrow Zn_3(PO_4)_2(s)$

Explanation:

Complete ionic equation : In complete ionic equation, all the substances that are strong electrolyte are present in an aqueous state as ions.

Net ionic equation : In the net ionic equations, we are not include the spectator ions in the equations.

Spectator ions : The ions present on reactant and product side which do not participate in a reactions. The same ions present on both the sides.

When sodium phosphate and zinc acetate then it gives zinc phosphate and sodium acetate as product.

The balanced molecular equation will be,

$2Na_3PO_4(aq)+3(CH_3COO)_2Zn(aq)\rightarrow 6CH_3COONa(aq)+Zn_3(PO_4)_2(s)$

The total ionic equation in separated aqueous solution will be,

$6Na^+(aq)+2PO_4^{3-}(aq)+6CH_3COO^-(aq)+3Zn^{2+}(aq)\rightarrow 6CH_3COO^-(aq)+6Na^+(aq)+Zn_3(PO_4)_2(s)$

In this equation, and are the spectator ions.

By removing the spectator ions from the balanced ionic equation, we get the net ionic equation.

The net ionic equation will be,

$2PO_4^{3-}(aq)+3Zn^{2+}(aq)\rightarrow Zn_3(PO_4)_2(s)$

5 0

3 years ago