What volume of 6.00 M HNO3 would be needed to completely react with 50.0g of Ca(OH)2 in the reaction below? Ca(OH)2(s) + 2HNO3(aq) ---> 2H2O(1) + Ca(NO3)2(aq)
1) calculate how many moles you have in 50.0g of Ca(OH)2 n=m/MM (number of moles equal to mass divided by the molecular mass) 2)By the stoichiometry of the reaction you can see that one mole of Ca(OH)2 will react with 2 moles of HNO3. Then you will need double amount of moles of HNO3 than moles of Ca(OH)2 calculated in the step 1. So, multiply the moles of Ca(OH)2 by 2 to find the moles of HNO3. 3) Calculate the volume of HNO3 M=n/V (molarity is equal to moles divided the volume in liters) V=n/M (divide the number of moles of HNO3 that you calculate in the step 2 by the concentration of the HNO3 given in the problem and you will have the volume in liters that you will need to react with the 50.0g of Ca(OH)2
For step 1) I got 1.48 moles. For step 2) I multiplied by 2 to get 2.96 For step 3) I am confused
For step 1) I got 1.48 moles. NO!! 50.0/74.1=0.67
Ca=54.09 O=16 H=1
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