consider the reaction.\nc(s)+o₂(g)→co₂(g)\nhow many grams of carbon should be burned in an excess of oxygen…

consider the reaction.\nc(s)+o₂(g)→co₂(g)\nhow many grams of carbon should be burned in an excess of oxygen at stp to obtain 2.21 l of carbon dioxide?\n1.18 g\n2.21 g\n4.12 g\n4.34 g

consider the reaction.\nc(s)+o₂(g)→co₂(g)\nhow many grams of carbon should be burned in an excess of oxygen at stp to obtain 2.21 l of carbon dioxide?\n1.18 g\n2.21 g\n4.12 g\n4.34 g

Answer

Explanation:

Step1: Calculate moles of CO₂

At STP, 1 mole of any gas occupies 22.4 L. So, the number of moles of CO₂, $n_{CO_2}=\frac{V}{V_m}$, where $V = 2.21$ L and $V_m=22.4$ L/mol. $n_{CO_2}=\frac{2.21}{22.4}\text{ mol}$

Step2: Determine moles of C

From the balanced chemical equation $C(s)+O_2(g)\rightarrow CO_2(g)$, the mole - ratio of C to CO₂ is 1:1. So, $n_{C}=n_{CO_2}=\frac{2.21}{22.4}\text{ mol}$

Step3: Calculate mass of C

The molar mass of C is $M_{C}=12.01$ g/mol. The mass of C, $m = n\times M$. So, $m_{C}=n_{C}\times M_{C}=\frac{2.21}{22.4}\text{ mol}\times12.01\text{ g/mol}$ $m_{C}=\frac{2.21\times12.01}{22.4}\text{ g}\approx1.18\text{ g}$

Answer:

  1. 1.18 g