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Nady [450]
4 years ago
7

__ P4(s) + __ O2(g) → __ P4O10(s) To balance this reaction equation, first write correct formulas for the reactants and products

, leaving the stoichiometric coefficients blank. We will balance P first. How many P4 molecules are needed to balance P for each P4O10?
Chemistry
1 answer:
Vedmedyk [2.9K]4 years ago
8 0

Answer:

P4(s) + 5 O2 (g)→ P4O10

Explanation:

If we desire to write a balanced chemical reaction equation, the rule of thumb is simple; the number of atoms of each element on the right hand side of the reaction equation must be the same as the number of atoms of the same element on the left hand side of the reaction equation. Once this condition is satisfied, the reaction equation is said to be balanced.

As we can see, we need one mole of P4 and five moles of O2 to produce one mole of P4O10.

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What are the elements in group 3-12 called
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Transition Metals

Explanation:

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How many grams of neutral salt will be obtained in the reaction of calcium oxide with 200 cm 3 of phosphoric acid solution whose
katen-ka-za [31]

Answer:

9.3 g of Ca3(PO4)2

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

3CaO + 2H3PO4 —> Ca3(PO4)2 + 3H2O

Next, we shall determine the number of mole of H3PO4 present in 200 cm³ of 0.3 mol/dm³ phosphoric acid (H3PO4) solution. This can be obtained as follow:

Molarity of H3PO4 = 0.3 mol/dm³

Volume = 200 cm³ = 200 cm³/1000 = 0.2 dm³

Mole of H3PO4 =?

Molarity = mole /Volume

0.3 = mole of H3PO4 /0.2

Cross multiply

Mole of H3PO4 = 0.3 × 0.2

Mole of H3PO4 = 0.06 mole

Next, we shall determine the number of mole of the salt, Ca3(PO4)2, obtained from the reaction. This can be obtained as shown below:

3CaO + 2H3PO4 —> Ca3(PO4)2 + 3H2O

From the balanced equation above,

2 moles of H3PO4 reacted to produced 1 mole of Ca3(PO4)2.

Therefore, 0.06 moles of H3PO4 will react to produce = (0.06 × 1)/2 = 0.03 mole of Ca3(PO4)2.

Thus, 0.03 mole of Ca3(PO4)2 is produced from the reaction.

Finally, we shall determine the mass of Ca3(PO4)2 produced as follow:

Mole of Ca3(PO4)2 = 0.03 mole

Molar mass of Ca3(PO4)2 = (40×3) + 2[31 + (16×4)]

= 120 + 2[31 + 64]

= 120 + 2[95]

= 120 + 190

= 310 g/mol

Mass of Ca3(PO4)2 =?

Mole = mass /Molar mass

0.03 = mass of Ca3(PO4)2 / 310

Cross multiply

Mass of Ca3(PO4)2 = 0.03 × 310

Mass of Ca3(PO4)2 = 9.3 g

Thus, 9.3 g of Ca3(PO4)2 was obtained from the reaction.

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3 years ago
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