?$解：根據(jù)化學(xué)方程式2\mathrm {H}_2+\mathrm {O}_2\xlongequal[]{點燃}2\mathrm {H}_2\mathrm {O}可知，$?
?$各物質(zhì)間的質(zhì)量之比為2×(1×2\mathrm {)}∶16×2∶2×(1×2+16\mathrm {)}=1∶8∶9$
?$則50\mathrm {g}氧氣只能支持?50\mathrm {g}×\frac { 1 }{ 8 }=6.25\mathrm {g}的氫氣燃燒，剩余10\mathrm {g}-6.25\mathrm {g}=3.75\mathrm {g}$?
?$反應(yīng)完畢生成水的質(zhì)量為6.25\mathrm {g}×9=56.25\mathrm {g}$?
