5a). Sodium reacts with excess oxygen to form sodium peroxide, Na,0,. 2Na + 02 → Na,02 Calculate the maximum mass of sodium peroxide formed when 4.60 g of sodium is burnt in excess oxygen. (A, values: Na = 23.0, 0 = 16.0)

5b). Tin oxide (IV) is reduced to tin by carbon. Carbon monoxide is also formed. Sno2 + 20 > Sn + 2C0 Calculate the mass of carbon that exactly reacts with 14.0 g of tin(IV) oxide. Give your answer to 3 significant figures. (A, values: C = 12.0,0 = 16.0, Sn =118.7)

5a) The maximum mass of sodium peroxide formed when 4.60 g of sodium is burnt in excess oxygen is; 7.8g of Sodium peroxide

5b) The mass of carbon that exactly reacts with 14.0g of tin(IV) oxide is; 2.23g of Carbon.

For questions (5a)

Sodium reacts with excess oxygen to form Sodium peroxide as follows;

2Na + 02 → Na202.

Molar mass of sodium and sodium peroxide are 23 and 78 respectively.

Since 2 units of sodium produces one unit of sodium peroxide.

Therefore,

46g of sodium = 78g of Na2O2.

4.60g of sodium = xg of Na2O2.

x = (78 × 4.6)/46

x = 7.8g of Na2O2

For question (5b)

Tin oxide (IV) is reduced to tin by carbon. Carbon monoxide is also formed as follows;

Sno2 + 2C > Sn + 2C0

From stoichiometry,

24g of Carbon = 150.7g of tin(IV) oxide.

xg of carbon = 14g of tin(IV) oxide

By Cross-product,

x = (14× 24)/150.7

x = 2.23g of carbon

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