Aluminum carbide (A l4 C3) liberates methane on treatment with water. The grams of aluminum carbide required to produce text11 text.2 textL of methane under STP conditions is [Given textAl text= text27 ]

Question

Aluminum carbide (A l4 C3) liberates methane on treatment with water. The grams of aluminum carbide required to produce text11 text.2 textL of methane under STP conditions is [Given textAl text= text27 ] (A) 48 (B) 72 (C) 144 (D) 24

Tardigrade Admin · Accepted Answer

Al4C3+12H2O arrow 3CH4+4Al(O H3) So text67 text.2 textL CH4 evolved by text144 textg Al4C3 text11 text.2 textL CH4 will be evolved = (144/67.2) × 11.2 = 24 textg

Q. Aluminum carbide $(A l_{4} C_{3})$ liberates methane on treatment with water. The grams of aluminum carbide required to produce $11 .2 L$ of methane under STP conditions is [Given $Al = 27$ ]

48

72

144

24

$A l_{4} C_{3} + 12 H_{2} O \to 3 C H_{4} + 4 A l (O H_{3})$

So $67 .2 L$ $C H_{4}$ evolved by $144 g$ $A l_{4} C_{3}$

$11 .2 L$ $C H_{4}$ will be evolved $= \frac{144}{67.2} \times 11.2 = 24 g$

Q. Aluminum carbide (Al4​C3​) liberates methane on treatment with water. The grams of aluminum carbide required to produce 11.2L of methane under STP conditions is [Given Al=27 ]

48

72

144

24

Al4​C3​+12H2​O→3CH4​+4Al(OH3​) So 67.2L CH4​ evolved by 144g Al4​C3​ 11.2L CH4​ will be evolved =67.2144​×11.2=24g

Q. Aluminum carbide $(A l_{4} C_{3})$ liberates methane on treatment with water. The grams of aluminum carbide required to produce $11 .2 L$ of methane under STP conditions is [Given $Al = 27$ ]

$A l_{4} C_{3} + 12 H_{2} O \to 3 C H_{4} + 4 A l (O H_{3})$

So $67 .2 L$ $C H_{4}$ evolved by $144 g$ $A l_{4} C_{3}$

$11 .2 L$ $C H_{4}$ will be evolved $= \frac{144}{67.2} \times 11.2 = 24 g$