JSP编码，新建了一个servlet，然后在web.xml中写入了配置信息，但是启动Tomcat时候为什么会失败？

web.xml内容
<?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns="http://java.sun.com/xml/ns/javaee" xmlns:web="http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd" xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_3_0.xsd" id="WebApp_ID" version="3.0">
<display-name>WebServlet</display-name>
<welcome-file-list>
<welcome-file>index.html</welcome-file>
<welcome-file>index.htm</welcome-file>
<welcome-file>index.jsp</welcome-file>
<welcome-file>default.html</welcome-file>
<welcome-file>default.htm</welcome-file>
<welcome-file>default.jsp</welcome-file>
</welcome-file-list>
<servlet>
<description>Hello</description>
<display-name>Hello</display-name>
<servlet-name>Hello</servlet-name>
<servlet-class>com.Hello</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>Hello</servlet-name>
<url-pattern>/Hello</url-pattern>
</servlet-mapping>
</web-app>
Servlet内Hello.java中只写了public void service(ServletRequest request, ServletResponse response) ;