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If 54.3 grams of tin (IV) chloride reacts with 23 L of sodium at STP, according to the following reaction, what mass of tin would be produced?
SnCla + 4Na -> 4NaCI + Sn

Respuesta :

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n SnCl₄ = m SnCl₄ / Mr SnCl₄

n SnCl₄ = 54.3 / 260.5

n SnCl₄ = 0.208 mol

n Na = V / 22.4 (at STP)

n Na = 23 / 22.4

n Na = 1.027 mol

Mole of Sn produced :

n Sn = (1/1) • 0.208

n Sn = 0.208 mol

m Sn = n Sn • Mr Sn

m Sn = 0.208 • 118.7

m Sn = 24.69 gr

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