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balandron [24]
3 years ago
14

How many grams of H3PO4 are produced when 10.0 moles of water react with an excess of P4O10?

Chemistry
2 answers:
vlada-n [284]3 years ago
8 0

The answer is 653 g

Pavel [41]3 years ago
8 0

Answer:

653.33gH_3PO_4

Explanation:

Hello,

In this case, by considering the given chemical reaction and the molar mass for phosphoric acid and the 4 to 6 relationship between phosphoric acid and water respectively, the required grams are result:

10.0molH_2O*\frac{4molH_3PO_4}{6molH_2O} *\frac{98gH_3PO_4}{1molH_3PO_4} =653.33gH_3PO_4

Best regards.

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Determine the mass of SO₂ that contains 6.075 × 10^26 S atoms.​
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\frac{6.075 \times 10^{26}}{6.022 \times 10^{23}}=1008.8010627 \text{ mol}

  • The atomic mass of sulfur is 32.06 amu.
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Answer:

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1. CuSO₄ (aq) + 2 KOH (aq) ----> Cu(OH)₂ (s) + K₂SO₄ (aq)

2. Ba(NO₃)₂ (aq) + K₂SO₄ (aq) + BaSO₄ (s) + 2 KNO₃ (aq)

The complete ionic equations are:

1. Ag + (aq) + NO₃- (aq) + I- (aq) + Na (aq) ---> AgI (s) + No₃- (aq) + Na+ (aq)

2. Cu²+ + SO₄²- (aq) + 2 K+ (aq) + 2 OH- (aq) ---> Cu(OH)₂ (s) + 2K+ (aq) + SO₄²- (aq)

The net ionic equations are:

1. Ca²+ (aq) + SO₄²- (aq) ---> CaSO₄ (s)

2. Ba²+ (aq) +SO₄²- (aq) ---> BaSO₄ (s)

Explanation:

A molecular equation is a balanced chemical equation which shows the reacting species as molecules rather than as componenet ions in their compounds with subscripts written beside the molecules to indicate the state in which they occur in the chemical reaction.

An ionic equation expresses the reacting species as components ions in a chemical reation. All the ions and molecules reacting are shown.

In a net ionic equation, the ions which remain in the ionic state also known as spectator ions are not written as part of the equation.

From the given attachment;

The molecular equations are:

1. CuSO₄ (aq) + 2 KOH (aq) ----> Cu(OH)₂ (s) + K₂SO₄ (aq)

2. Ba(NO₃)₂ (aq) + K₂SO₄ (aq) + BaSO₄ (s) + 2 KNO₃ (aq)

The complete ionic equations are:

1. Ag + (aq) + NO₃- (aq) + I- (aq) + Na (aq) ---> AgI (s) + No₃- (aq) + Na+ (aq)

2. Cu²+ + SO₄²- (aq) + 2 K+ (aq) + 2 OH- (aq) ---> Cu(OH)₂ (s) + 2K+ (aq) + SO₄²- (aq)

The net ionic equations are:

1. Ca²+ (aq) + SO₄²- (aq) ---> CaSO₄ (s)

2. Ba²+ (aq) +SO₄²- (aq) ---> BaSO₄ (s)

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