If 87.1 g of solid CaCO3 and 690 mL of 0.870 M aqueous H2SO4 are reacted stoichiometrically according to the b

Can someone guide me through these problems and help me?

If 420 mL of 1.07 M aqueous HClO4 and 770 mL of 0.234 M aqueous Ba(OH)2 are reacted stoichiometrically according to the balanced equation, how many milliliters of 1.07 M aqueous HClO4 remain? Round your answer to 3 significant figures.

Ba(OH)2(aq) + 2HClO4(aq) → Ba(ClO4)2(aq) + 2H2O(l)


If 12.1 L of gaseous Cl2 measured at STP reacts stoichiometrically according to the balanced equation, how many milliliters of 3.59 M aqueous NaBr are required?

2NaBr(aq) + Cl2(g) → 2NaCl(aq) + Br2(l)


If 820 mL of 0.366 M aqueous LiOH and 23.8 g of gaseous CO2 are reacted stoichiometrically according to the balanced equation, how many grams of gaseous CO2 remain? Round your answer to 3 significant figures.


If 240 mL of 3.21 M aqueous BaCl2 and 340 mL of 2.12 M aqueous Al2(SO4)3 are reacted stoichiometrically according to the balanced equation, how many milliliters of 2.12 M aqueous Al2(SO4)3 remain? Round your answer to 3 significant figures.

Al2(SO4)3(aq) + 3BaCl2(aq) → 2AlCl3(aq) + 3BaSO4(s)


If 160 mL of 4.4375 M aqueous Al2(SO4)3 reacts stoichiometrically according to the balanced equation, how many moles of solid BaSO4 are produced?

Al2(SO4)3(aq) + 3BaCl2(aq) → 2AlCl3(aq) + 3BaSO4(s)

If 860 mL of 0.72093 M aqueous CaCl2 reacts stoichiometrically according to the balanced equation, how many milliliters of 3.50 M aqueous NaCl are produced?

3CaCl2(aq) + 2Na3PO4(aq) → Ca3(PO4)2(s) + 6NaCl(aq)

can't do the first one since you cut off the question

2.
Ba(OH)2(aq) + 2HClO4(aq) → Ba(ClO4)2(aq) + 2H2O(l)
0.77L x 0.234M Ba(OH)2 = 0.18moles Ba(OH)2 present

0.18moles Ba(OH)2 x (2HClO4 / 1Ba(OH)2) = 0.36moles Ba(OH)2 used
for 0.36moles Ba(OH)2, this would be xL of 1.07M?
0.36 / xL = 1.07M
0.336L Ba(OH)2 would have been used of the 0.42L therefore, 0.0836L or 83.6ml Ba(OH)2 is left over.


3. 12.1L Cl2 at STP / 22.4L/mole = 0.54moles Cl2 present
0.54moles Cl2 x (2NaBr / 1Cl2) = 1.08moles NaBr required
1.08moles NaBr / 3.59M = 0.301L or 301ml

4. 23.8g CO2 / 44g/mole = moles CO2
0.92L x 0.366M LiOH = moles LiOH
LiOH is limiting so: moles CO2 required = moles LiOH x (ratio from balanced equation which is missing from the question

5 - 7 same as 2 - 4