Iron reacts with excess Sulphuric acid to give Iron II sulphate. The equation for the reaction is Fe s + H 2 SO 4 aq → FeSO 4 aq + H 2 g . 5 g of Iron gives 13 . 6 g of Iron II sulphate. Using excess acid, how much Iron II sulphate can be obtained from 10 g of Iron.

Fe s + H 2 SO 4 aq → FeSO 4 aq + H 2 g Atomic mass of Iron is 56 . Given that 5 g of Iron gives 13 . 6 g of Iron sulphate. Therefore, 10 g of Iron would give, = 10 × 13 . 6 5 = 27 . 2 g of Iron sulphate.

Iron reacts with excess Sulphuric acid to give Iron II sulphate. The equation for the reaction is Fe s + H 2 SO 4 aq → FeSO 4 aq + H 2 g . 5 g of Iron gives 13 . 6 g of Iron II sulphate. Using excess acid, how much Iron II sulphate can be obtained from 1 g of Iron?

Fe s + H 2 SO 4 aq → FeSO 4 aq + H 2 g Atomic mass of Iron is 56 . Given that, 5 g of Iron gives 13 . 6 g of Iron sulphate. Therefore, 1 g of Iron gives = 1 × 13 . 6 5 = 2 . 72 g of Iron sulphate.

Iron reacts with excess Sulphuric acid to give Iron II sulphate. The equation for the reaction is Fe s + H 2 SO 4 aq → FeSO 4 aq + H 2 g . 5 g of Iron gives 13 . 6 g of Iron II sulphate. How much Iron will be needed to make 136 g of Iron II sulphate?

Fe s + H 2 SO 4 aq → FeSO 4 aq + H 2 g Atomic mass of Iron is 56 . Given that, 5 g of Iron gives 13 . 6 g of Iron sulphate. Therefore, 136 g of Iron sulphate would be obtained from = 5 × 136 13 . 6 = 50 g of Iron.

Iron reacts with excess Sulphuric acid to give Iron II sulphate. The equation for the reaction is Fe s + H 2 SO 4 aq → FeSO 4 aq + H 2 g . 5 g of Iron gives 13 . 6 g of Iron II sulphate. A 10 g of sample of impure Iron II sulphate contains 8 g of iron II sulphate. Calculate the percentage purity of Iron II sulphate.

A pure substance has nothing else mixed with it. But substances always contain unwanted impurities. A 10 g of iron sulphate contains 8 g of pure iron sulphate. Hence, its percentage of purity is: % purity of substance = mass of pure substance in it total mass × 100 % = 8 10 × 100 % = 80 %

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Upper Secondary: IGCSE

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Calculate the relative formula mass for the ionic compound Sodium chloride, $NaCl$ .

Important Questions on Reacting Masses, and Chemical Equations

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Upper Secondary: IGCSE

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Calculate the relative formula mass for the ionic compound hydrated Iron

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HARD

Upper Secondary: IGCSE

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Complete Chemistry for Cambridge IGCSE® Second Edition>Chapter 5 - Reacting Masses, and Chemical Equations>Checkup on Chapter 5>Q 12

Iron reacts with excess Sulphuric acid to give Iron $(II)$ sulphate. The equation for the reaction is $Fe (s) + H_{2} {SO}_{4} (aq) \to {FeSO}_{4} (aq) + H_{2} (g)$ . $5 g$ of Iron gives $13.6 g$ of Iron $(II)$ sulphate.

Using excess acid, how much Iron $(II)$ sulphate can be obtained from $10 g$ of Iron.

HARD

Upper Secondary: IGCSE

IMPORTANT

Complete Chemistry for Cambridge IGCSE® Second Edition>Chapter 5 - Reacting Masses, and Chemical Equations>Checkup on Chapter 5>Q 12

Using excess acid, how much Iron $(II)$ sulphate can be obtained from $1 g$ of Iron?

HARD

Upper Secondary: IGCSE

IMPORTANT

Complete Chemistry for Cambridge IGCSE® Second Edition>Chapter 5 - Reacting Masses, and Chemical Equations>Checkup on Chapter 5>Q 12

Iron reacts with excess Sulphuric acid to give Iron $(II)$ sulphate. The equation for the reaction is $Fe (s) + H_{2} {SO}_{4} (aq) \to {FeSO}_{4} (aq) + H_{2} (g) . 5 g$ of Iron gives $13.6 g$ of Iron $(II)$ sulphate.

How much Iron will be needed to make $136 g$ of Iron $(II)$ sulphate?

HARD

Upper Secondary: IGCSE

IMPORTANT

Complete Chemistry for Cambridge IGCSE® Second Edition>Chapter 5 - Reacting Masses, and Chemical Equations>Checkup on Chapter 5>Q 12

A $10 g$ of sample of impure Iron $(II)$ sulphate contains $8 g$ of iron $(II)$ sulphate. Calculate the percentage purity of Iron $(II)$ sulphate.

Calculate the relative formula mass for the ionic compound Sodium chloride, NaCl.

Important Questions on Reacting Masses, and Chemical Equations

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Calculate the relative formula mass for the ionic compound Sodium chloride, $NaCl$ .