How many mL of 6.0M H2SO4 are needed to react with 0.80g of CuO according to CuO(s)+H2SO4(aq)->CuSO4(aq)+H2O?

Question

trying to do this for chemistry and I'm royally stuck!!1 Any help would be greatly appreciated!

- part of the question was obviously cut off... This is the second half of the equation...
-->CuSO4(aq) + H2O(l)

Answer 1

Convert the mass of CuO into moles by dividing 0.80 by the molecular weight of CuO. Now, from your equation you see that 1 mole of H2SO4 will react with 1 mole of CuO. Consequently, you will need the same number of moles of H2SO4 as you have CuO. If you divide that number of moles by 6 mol/L, that will give you the Liters of H2SO4 that you need. Multiply that by 1000 to convert it to mL, and you're done.

Answer 2

First convert 0.80g of CuO to moles utilising the molar mass of CuO. Then use the mole ratio from the balanced equation to calculate the variety of moles of H2SO4 mandatory. (You get fortunate right here using fact it extremely is a one million:one million ratio). understanding that 6M H2SO4 ability you have 6 moles according to liter, or according to one thousand mL, take the variety of moles of H2SO4 you obtain from step one million and multiply it via one thousand mL/6 moles H2SO4 and you have an answer in mLs.

Answer 3

n(CuO) = 0.80 / (64 + 16) = 0.01mol

therefore since 1:1 molar ratio of CuO:H2SO4
0.01mol of H2SO4 is required

V(H2SO4) = n(H2SO4) / ([H2SO4]* 10^-3)
V(H2SO4) = 0.01 / (6*10^-3)
V(H2SO4) = 1.66ml