Phosphorus can be prepared from calcium phosphate by the following reaction: 2 Ca 3 ( PO 4 ) 2 ( s ) + 6 SiO 2 ( s ) + 10 C ( s ) → 6 CaSiO 3 ( s ) + P 4 ( s ) + 10 CO ( g ) Phosphorite is a mineral that contains Ca 3 (PO 4 ) 2 plus other non-phosphorus-containing compounds. What is the maximum amount of P 4 that can be produced from 1.0 kg of phosphorite if the phorphorite sample is 75% Ca 3 (PO 4 ) 2 by mass? Assume an excess of the other reactants.

Q: Expert Answer

Interpretation Introduction Interpretation: The maximum amount of P 4 that can be produced from 1.0 kg of phosphorite is to be calculated. Concept introduction: The mass of a substance can be obtained by using the number of moles of the substance present and its molar mass. The formula used to calculate the mass of a given substance is, Mass of the substance = ( Number of moles ) × ( Molar mass of the substance ) To determine: The maximum amount of P 4 that can be made using 1.0 kg of phosphorite. Explanation Given The stated chemical equation is, 2 Ca 3 ( PO 4 ) 2 ( s ) + 6 SiO 2 ( s ) + 10 C ( s ) → 6 CaSiO 3 ( s ) + P 4 ( s ) + 10 CO ( g ) The given mass of phosphorite is 1.0 kg ( 1000 g ) . The mass of Ca 3 ( PO 4 ) 2 is 75 % of the mass of phosphorite. Therefore, the mass of Ca 3 ( PO 4 ) 2 = ( 75 100 × 1000 ) g = 750 g Molar mass of Ca 3 ( PO 4 ) 2 = 3Ca + 2 P + 8 O = ( ( 3 × 40 ) + ( 2 × 31 ) + ( 8 × 16 ) ) g / mol = 310 g / mol Formula The number of moles of a substance is calculated by the formula, Number of moles = Given mass of the substance Atomic mass of the substance Substitute the value of the given mass and the atomic mass of Ca 3 ( PO 4 ) 2 in the above expression. Number of moles of Ca 3 ( PO 4 ) 2 = 750 g 310 g / mol = 2

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Chemistry: An Atoms First Approach

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Steven S. Zumdahl + 1 other

Publisher: Cengage Learning

ISBN: 9781305079243

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Interpretation Introduction

Interpretation: The maximum amount of P4 that can be produced from 1.0 kg of phosphorite is to be calculated.

Concept introduction: The mass of a substance can be obtained by using the number of moles of the substance present and its molar mass. The formula used to calculate the mass of a given substance is,

Mass of the substance=(Number of moles)×(Molar mass of the substance)

To determine: The maximum amount of P4 that can be made using 1.0 kg of phosphorite.

Explanation of Solution

Given

The stated chemical equation is,

2Ca3(PO4)2(s)+6SiO2(s)+10C(s)→6CaSiO3(s)+P4(s)+10CO(g)

The given mass of phosphorite is 1.0 kg (1000 g) .

The mass of Ca3(PO4)2 is 75 % of the mass of phosphorite.

Therefore, the mass of Ca3(PO4)2 =(75100×1000) g=750 g

Molar mass of Ca3(PO4)2 =3Ca+2P+8O=((3×40)+(2×31)+(8×16)) g/mol=310 g/mol

Formula

The number of moles of a substance is calculated by the formula,

Number of moles=Given mass of the substanceAtomic mass of the substance

Substitute the value of the given mass and the atomic mass of Ca3(PO4)2 in the above expression.

Number of moles of Ca3(PO4)2=750 g310 g/mol=2

Chemistry: An Atoms First Approach

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Chemistry: An Atoms First Approach

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