Answer:
Cancel out CO because it appears as a reactant in one intermediate reaction and a product in the other intermediate reaction.
Explanation:
The CO appears twice hence in he intermediate reaction it only forms path of the enabling reagents and it further reacts to form the final product. Accounting for the CO in the intermediate reaction that undergoes further reaction will impact on the stoichiometry of the reaction.
Answer: 37.5grams of Cu(NO3)2
Cu(1mol) + 2HNO3(2mol) —> Cu(NO3)2 + H2
<em>125 grams of Cu(1mol) reacts with 75 grams of HNO3(2mol)</em>
<em><u>HNO3 is the limiting substance, therefore, 75 grams is the limiting quantity.</u></em>
<em>Therefore, 2mol of HNO3 forms 1mol of Cu(NO3)2</em>
<em>75 grams of HNO3 forms...75grams x 1mol/2mol = 37.5 grams of Cu(NO3)2</em>
Answer:
if a 40.0-gram sample of the gas occupies 11.2 liters of space at STP? A balloon is filled with 5 moles of helium gas.
Explanation:
