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1 g of Mg is burnt in a closed vessel containing 0.5 g of O_2. Which reactant is limiting reagent and how much of the excess reactant will be left?

Answer» <html><body><p></p>Solution :Carbon burns in oxygen according to the following <a href="https://interviewquestions.tuteehub.com/tag/equation-974081" style="font-weight:bold;" target="_blank" title="Click to know more about EQUATION">EQUATION</a>. <br/> `underset(12.01 <a href="https://interviewquestions.tuteehub.com/tag/g-1003017" style="font-weight:bold;" target="_blank" title="Click to know more about G">G</a>)C + underset(32.0 g)(O_(2)) to underset(12.01 + 32.0 = 44.01 g)(CO_(2))` <br/> `therefore` 12.01 g of carbon require for combustion, = 32.0 g <br/> `therefore 1.0` g of carbon will require `=(32.0)/(12.01) xx 1.0 = 2.7 g` <br/> Hence, the oxygen present in the closed vessel is much <a href="https://interviewquestions.tuteehub.com/tag/less-1071906" style="font-weight:bold;" target="_blank" title="Click to know more about LESS">LESS</a> than the required amount. The amount of `O_(2)`will thus be completely consumed and carbon will be left in excess. Thus, `O_2` is the <a href="https://interviewquestions.tuteehub.com/tag/limiting-2791961" style="font-weight:bold;" target="_blank" title="Click to know more about LIMITING">LIMITING</a> reagent in this case. <br/> `therefore 32.0` g of oxygen burn carbon = 12.01 g <br/> 1.5 g of oxygen will burn carbon `=12.01/32.0 xx 1.5 = 0.56 g` <br/> Hence, carbon left unreacted = `1.0 - 0.56 = 0.44` g Now, `therefore32.0` g of oxygen combine with carbon to <a href="https://interviewquestions.tuteehub.com/tag/give-468520" style="font-weight:bold;" target="_blank" title="Click to know more about GIVE">GIVE</a> `CO_2 = 44.01` g <br/> `therefore` 1.5 g of oxygen will give `CO_(2) = (44.01)/(32.0) xx 1.5 = 2.1 g` <br/> Hence, carbon is left in excess in the given reaction. The amount of excess reactant is 0.44 g and the mass of the product `CO_2` formed in the reaction is 2.1 g</body></html>


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