$解：設(shè)25g該電石樣品中CaC_{2}的質(zhì)量為x。$
$CaC_{2}+2H_{2}O = Ca(OH)_{2}+C_{2}H_{2}\uparrow$
$64 ~~~~~~~~~~~26$
$x ~~~~~~~~~~~7.8g$
$\dfrac{64}{26}=\dfrac{x}{7.8g}$
$x = \dfrac{64×7.8g}{26}=19.2g$
$該電石樣品中CaC_{2}的質(zhì)量分數(shù)為\dfrac{19.2g}{25g}×100\% = 76.8\%$
$答：該電石樣品中CaC_{2}的質(zhì)量分數(shù)為76.8\%。$
