How many mL of 6.0M H2SO4 are required to react with 0.80g of CuO

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To determine the volume of 6.0M H2SO4 required to react with 0.80g of CuO, one must first calculate the moles of CuO using its molar mass and then apply stoichiometric principles based on the balanced reaction CuO + H2SO4 → CuSO4 + H2O. The stoichiometry indicates a 1:1 molar ratio between CuO and H2SO4. Additionally, for the reaction of 2.00g of Zn with 1.75g of CuSO4, the remaining grams of Zn can be found by identifying the limiting reactant and calculating the theoretical yield of the products. Understanding limiting and excess reactants is crucial for these calculations.

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How many mL of 6.0M H2SO4 are required to react with 0.80g of CuO according to CuO + H2SO4---->CuSO4 + H2O?



If 2.00 g of Zn is allowed to react with 1.75 g of CuSO4 according to CuSO4 + Zn---->ZnSO4 + Cu, how many grams of Zn will remain after the reaction is complete?


If anyone could help with either of these questions, it would be greatly appreiated. Thanks!
 
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Hint: find limiting and excess reactants. the limiting reactant is your theoretical yield
 

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