D - density: 40 mg/ml
V - volume: 175 ml
m - mass: ??
__________
d = m/V
m = V×d
m = 175×40
m = 7000 mg = 7 g
:•)
Answer:
1.17 grams of HCl can neutralize 2.7 grams sodium bicarbonate
Explanation:
Step 1: Data given
Mass of sodium bicarbonate = 2.7 grams
Step 2: The balanced equation
HCl + NaHCO3 ⇔ NaCl + H2O + CO2
Step 3: Calculate moles NaHCO3
moles NaHCO3 =2.7 g / 84 g/mol= 0.032 moles
Step 4: Calculate moles HCl
For 1 mol NaHCO3 we need 1 mol HCl
For 0.032 moles NaHCO3 = 0.032 moles HCl
Step 5: Calculate mass HCl
Mass HCl = moles HCl * molar mass HCl
mass HCl = 0.032 * 36.46 g/mol= 1.17 grams
1.17 grams of HCl can neutralize 2.7 grams sodium bicarbonate
6.069 grams is the mass of hydrogen formed when 27 g of aluminum reacts with excess hydrochloric acid.
Explanation:
Balanced equation for the reaction:
2 Al + 6 HCl → 2 AlCl₃ + 3 H₂
data given:
mass of aluminum = 27 grams
atomic mass of one mole of aluminum = 26.89 grams/mole
formula to calculate number of moles:
number of moles = 
number of moles = 
= 1.004 moles of aluminum will react
from the balanced equation:
2 moles of Al reacted to form 3 moles of H2
1.004 moles of Al will produce x moles of H2
= 
x = 3.012 moles of H2 will be formed.
mass will be calculated as number of moles multiplied by atomic weight
mass of 3.012 moles of hydrogen ?(atomic weight of one mole H2 = 2.015 grams)
= 3.012 x 2.015
= 6.069 grams of H2 will be formed.
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