Question

1. determine the chemical amount of silver produced when 118 g of zinc reacts completely with excess silver nitrate, as shown in the following chemical reaction: zn(s) + 2agno₃(aq) → 2ag(s) + zn(no₃)₂(aq) 3.61 moles 195 moles 0.902 moles 1.80 moles

Explanation:

Step1: Calculate moles of zinc

First, find the molar - mass of zinc ($Zn$). The molar mass of $Zn$ is $M_{Zn}=65.38\ g/mol$. Use the formula $n=\frac{m}{M}$, where $m = 118\ g$ and $M = 65.38\ g/mol$. So, $n_{Zn}=\frac{118\ g}{65.38\ g/mol}\approx1.80\ mol$.

Step2: Use mole - ratio from the balanced equation

From the balanced chemical equation $Zn(s)+2AgNO_3(aq)
ightarrow2Ag(s)+Zn(NO_3)_2(aq)$, the mole - ratio of $Zn$ to $Ag$ is $n_{Ag}/n_{Zn}=2/1$.
Since $n_{Zn} = 1.80\ mol$, then $n_{Ag}=2\times n_{Zn}$.
$n_{Ag}=2\times1.80\ mol = 3.60\ mol\approx3.61\ mol$.

Answer:

3.61 moles