Question

Calculați masa de C3H8 care consuma la ardere un volum de 4480l in condiții normale​

Answer 1

miu.C3H8 = 3x12+8x1 = 44 g/mol

stim ca niu = m/miu = V/Vm

=> m = Vxmiu/Vm = 4480x44/22,4 = 8800 g C3H8

Answer 2

$V=4 \ 480 \ l \\ M(C_3H_8)=12 \cdot 3+1 \cdot 8=36+8=44 \ g/mol; \\ V_m=22,4 \ l/mol \\ \nu=\frac{m}{M}=\frac{V}{V_m} \ \Rightarrow m=\frac{M \cdot V}{V_m} \\ m=\frac{44 \ g/mol \cdot 4 \ 480 \ l}{22,4 \ l/mol}=8 \ 800 \ g \\ \Rightarrow \boxed{m(C_3H_8)=8,8 \ kg}$