The amount of zinc required to produce 224 $$mL$$ of $${H_2}$$ at $$STP$$ on treatment with $$dil.$$ $${H_2}S{O_4}$$ will be
A.
6.5$$\,g$$
B.
0.65$$\,g$$
C.
65$$\,g$$
D.
0.065$$\,g$$
Answer :
0.65$$\,g$$
Solution :
$$Zn + {H_2}S{O_4} \to ZnS{O_4} + {H_2}$$
$$65\,g\,\,Zn$$ gives $$1$$ $$mole$$ of $${H_2} = 22400\,mL$$ of $${H_2}\,224\,mL$$ of $${H_2}$$ will be obtained from $$0.65\,g\,Zn.$$
Releted MCQ Question on Physical Chemistry >> Some Basic Concepts in Chemistry
Releted Question 1
$$27 g$$ of $$Al$$ will react completely with how many grams of oxygen?